Triangles

(i)

Both congruent and similar because

(a) all squares have same shape and size

(b) their corresponding angles are equal

(c) their corresponding sides are proportional

(ii)

Both congruent and similar because all circles have same shape and size

(iii) Similar because all rhombuses have the same angle, but size can vary.

(iv) Similar because all photographs have the same shape but not necessarily the same size.

(v) Two polygons having equal numbers of sides are similar if their corresponding angles are equal and their corresponding sides are Proportional

(i) This statement is false because all the congruent figures are similar, but similar figures need not be congruent.

E.g. Two equilateral triangles having sides 1cm and 2cm.

In case of equilateral triangles, all the sides are equal, and all the angles are of 60°.

But here, their corresponding angles are equal, but sides of triangle ABC and PQR are not equal in length.

So, they are similar figures but not congruent.

(ii) This statement is true because all congruent figures are similar, but similar figures need not be congruent.

(iii) This statement is false because for two triangles to be similar to the angles in one triangle must have the same values as the angles in the other triangle. The sides must be proportionate.

E.g.

These are the two isosceles triangles having two equal sides, but we can see that the sides are not proportionate.

(iv) This statement is false.

Suppose these are two right-angled triangles

Here, both of the triangles are right-angled, but other corresponding two angles are not equal. So, these are not similar figures.

(v) This statement is true because all the angles in a square are right angles and all the sides are equal. Hence, a smaller square can be enlarged to the size of a larger square, and vice-versa is also true.

(vi) This statement is false because similarity preserves the ratio of length. Therefore, two rectangles with a different ratio between their sides cannot be similar.

(vii) This statement is true because photographs are produced by projecting the image from a negative through an enlarger to a photographic paper. The enlarger reproduces the image from the negative but makes it bigger. The images are not identical and are not of the same size, but they are similar.

(viii) This statement is false because here the photograph of a person is taken at the different ages.

(i) (a) Two circles having radii 2cm and different centres

In this, both of them have the same radii, but their centres are different.

(b) Two squares having the same length of side 5cm

We know that in a square all the sides are equal and all angles are of 90°. So, these two squares are congruent.

(ii) (a) Two equilateral triangles having sides 1cm and 2cm.

In case of equilateral triangles, all the sides are equal, and all the angles are of 60°.

But here, their corresponding angles are equal, but sides of triangle ABC and PQR are not equal in length.

So, they are similar figures but not congruent.

(b) Two circles having radii 1cm and 2cm

Both of the figures are of circle but they are having different radii. So, they are similar but not congruent.

(iii) (a) A rhombus and a rectangle

In the case of a rhombus, all the sides are equal, and the angles can either be right angles or combination of acute and obtuse angles but in rectangle all angles are equal, and opposite sides are equal.

Hence, a rhombus and a rectangle are non-similar figures.

(b)

Here, both of the triangles are right-angled but other two angles are not equal. So, these are not similar figures.

Two polygons of a same number of sides are similar, if

a) all the corresponding angles are equal.

b) all the corresponding sides are in the same ratio (or proportion)

In case of right-angled triangle PQR and ABC

and

The corresponding sides of a right-angled triangle ABC and PQR are proportional, and their corresponding angles are not equal. Hence, triangles ABC and PQR are not similar.

Two polygons of the same number of sides are similar, if

a) All the corresponding angles are equal.

b) all the corresponding sides are in the same ratio (or proportion)

In case of rectangles ABCD and PQRS

And it is given that both are rectangles and we know that, in rectangle all angles are of 90°

The corresponding sides of a rectangle ABCD and PQRS are proportional, and their corresponding angles are equal. Hence, rectangles ABCD and PQRS are similar.

Two polygons of the same number of sides are similar, if

a) All the corresponding angles are equal.

b) all the corresponding sides are in the same ratio (or proportion)

In case of quadrilaterals ABCD and PQRS

and ∠A=∠B =∠C =∠D =∠P =∠Q =∠R=∠S =90°

The corresponding sides of a quadrilateral ABCD and PQRS are not proportional. Hence, quadrilaterals ABCD and PQRS are not similar.

Two polygons of a same number of sides are similar, if

a) all the corresponding angles are equal.

b) all the corresponding sides are in the same ratio (or proportion)

In case of polygons ABCD and PQRS

and ∠A=∠B =∠C =∠D =90° but ∠P ,∠Q ,∠R,∠S≠90°

The corresponding sides of a polygon ABCD and PQRS are proportional, but their corresponding angles are not equal. Hence, polygon ABCD and PQRS are not similar.

Two polygons of a same number of sides are similar if

a) all the corresponding angles are equal.

b) all the corresponding sides are in the same ratio (or proportion)

In case of polygons ABCD and PQRS

and ∠A=∠P =105°, ∠B =∠Q =100°, ∠C =∠R=70°, ∠D=∠S =85°

The corresponding sides of a polygon ABCD and PQRS are proportional, and their corresponding angles are also equal. Hence, polygon ABCD and PQRS are similar.

Two polygons of the same number of sides are similar, if

a) All the corresponding angles are equal.

b) all the corresponding sides are in the same ratio (or proportion)

In case of polygons ABCD and PQRS

Clearly, the corresponding sides of a polygon ABCD and PQRS are not proportional. Hence, polygon ABCD and PQRS are not similar.

Two polygons of the same number of sides are similar if

a) all the corresponding angles are equal.

b) all the corresponding sides are in the same ratio (or proportion)

In case of polygons □ ABCD and ◊ ABCD

The corresponding sides of a polygon ABCD and ABCD are proportional, but their corresponding angles are not equal as we can see the first figure is of a square (all angles are of 90°) and other is of a rhombus (in rhombus the diagonal meet in the middle at a right angle). Hence, polygon ABCD and ABCD are not similar.

Given: PQ || BC

and

By Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

(i)

Given: DE || BC

[by basic proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.]

[given: AD= 1.5cm, DB =3cm & AE =1cm]

⇒ EC = 2cm

(ii)

Given: DB = 7.2 cm, AE = 1.8 cm and EC = 5.4 cm

and DE || BC

[by basic proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.]

⇒ AD = 2.4cm

(i) Given: DE || BC

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

[by basic proportionality theorem]

(ii) Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

By basic proportionality theorem, we know that

(iii) From part (i), we know that

On adding 1 to both the sides, we get

(iv) From part (iii), we have

Given: DE || BC

and

To find: AE: EC

Given: DE || BC

BasicProportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

[by basic proportionality theorem]

Given: AD = x

DB =16, AE = 34 and EC = 17

Given: DE || BC

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

So, by basic proportionality theorem

⇒ x = 32

Given: AD = 2cm, DB =3cm, AE = 5cm

and DE || BC

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

So, by basic proportionality theorem

⇒ EC = 7.5 cm

Given: AD = 2.4cm, AE = 3.2cm and EC = 4.8cm

and DE || BC

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

So, by basic proportionality theorem

⇒ DB = 3.6 cm

or BD = 3.6 cm

Given: DE || BC

and

To find :

Given: DE || BC

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

[by basic proportionality theorem]

Given: AP = 8cm , AB =12cm, AQ = (3x)cm and QC = (x+2)cm

and PQ || BC

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

[by basic proportionality theorem]

⇒ 2(x+2) = 3x

⇒ 2x + 4 = 3x

⇒ 2x – 3x = - 4

⇒ x = 4

Given: AD = 4, DB =x – 4, AE = x – 3 and EC = 3x – 19

and DE || BC

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

So, by basic proportionality theorem

⇒ 4(3x – 19) = (x – 4)(x – 3)

⇒ 12x – 76 = x² – 3x – 4x + 12

⇒ 12x – 76 = x² –7x + 12

⇒ x² –7x + 12 – 12x + 76 = 0

⇒ x² –19x + 88 = 0

Solving the Quadratic equation by splitting themiddle term, we get,

⇒ x² –11x – 8x + 88 = 0

⇒ x(x – 11) – 8(x – 11) = 0

⇒ (x – 8)(x – 11) =0

⇒ x = 8 and 11

Given: AB = 12cm, AD =8cm, AE = 12cm and AC = 18cm

To Prove: DE || BC

In ABC,

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

Hence, DE || BC [by converse of basic proportionality theorem]

Hence, Proved.

(i) Given: AP= 8 cm, PB = 3 cm, AC = 22 cm and AQ =16 cm

To find: PQ || BC

In ABC,

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

Hence, PQ || BC [by converse of basic proportionality theorem]

Hence, Proved.

(ii) Given: AB= 1.28 cm, AC = 2.56 cm, AP= 0.16 cm and AQ = 0.32 cm

To find: PQ || BC

In ABC,

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

Hence, PQ || BC [by converse of basic proportionality theorem]

Hence, Proved.

(iii) Given: AB = 5 cm, AC =10 cm, AP= 4 cm, AQ = 8 cm

To find: PQ || BC

In ABC,

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

Hence, PQ || BC [by converse of basic proportionality theorem]

Hence, Proved.

(iv) Given: AP= 4 cm, PB= 4.5 cm, AQ = 4 cm, QC = 5 cm

To find: PQ || BC

In ABC,

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

⇒ PQ is not parallel to BC

Given: AD is the bisector of BAC

and by Angle-Bisector theorem which states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides.

⇒ AB = 3.6cm

Given: AD is the bisector of BAC

and by Angle-Bisector theorem which states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides.

And BC – CD = DB

⇒ 12 – 4.5 = DB

⇒ DB = 7.5cm

Given: AD is the bisector of A

and by Angle-Bisector theorem which states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides.

⇒ BD = 2cm

(i) In ABC,

∠A = 70° and ∠B = 50°

And we know that, sum of the angles = 180°

⇒∠A + ∠B +∠C = 180°

⇒70° + 50° +∠C = 180°

⇒ 120° +∠C = 180°

⇒ ∠C = 60°

And In DEF

∠F = 70° and ∠E = 50°

And we know that, sum of the angles = 180°

⇒∠D + ∠E +∠F = 180°

⇒∠D + 50° + 70° = 180°

⇒ ∠D = 60°

Yes,ABC ~ DEF [by AAA similarity criterion]

(ii) In ABC and PQR

Here,

As,

So, ABC ~ PQR [by SSS similarity criterion]

(iii) InMNL and PQR

∠NML = ∠PQR = 70°

and

No, the two triangles are not similar.

(iv) InPQR and LMN

∠PQR = ∠LNM = 50°

and

∴PQR ~ LMN [by SAS similarity criterion]

(v) In LMP and DEF

Here,

As

So, no two triangles are not similar

(vi) In ABC and PQR

∠A = ∠Q = 85°

∠B = ∠P = 60°

and ∠C =∠R = 35°

So, PQR ~ LMN [by AAA similarity]

(vii) In ABC and PQR

Here,

As,

So, ABC ~ PQR [by SSS similarity criterion]

Given: ABCD is a trapezium with AB || CD

and diagonals AB and CD intersecting at O

To find: y

Firstly, we prove that OAB ~ ODC

Let OAB and ODC

∠AOB = ∠COD [vertically opposite angles]

∠OBA = ∠ODC [∵AB || CD with BD as transversal.

alternate angles are equal]

∠OAB = ∠OCD [∵AB || CD with BD as transversal.

alternate angles are equal]

∴ OAB ~ ODC [by AAA similarity]

Since triangles are similar. Hence corresponding sides are proportional.

⇒ y = 6cm

Given: PQ || BC

To find: AP and AQ

Since, PQ || BC, AB is transversal, then,

APQ = ABC [by corresponding angles]

Since, PQ || BC, AC is transversal, then,

APQ = ABC [by corresponding angles]

In APQ andABC

∠APQ = ∠ABC

∠AQP = ∠ACB

∴ APQ ≅ ABC [by AAA similarity]

Since, the corresponding sides of similar triangles are proportional

⇒AP = 2.3

Now, taking

⇒AQ = 2.75

Therefore, AP = 2.3cm and AQ = 2.75cm

Given: ABR ~ PQR

As, ABR and PQR are similar

⇒BR = 70cm

and PR = AP – AR = 72 – 45 = 27cm

Let us take OAQ and OBP

∠AOQ = ∠BOP (vertically opposite angles)

∠OAQ = ∠OBP (each 90°)

∴ OAQ ~OBP (by AA similarity criterion)

Given: AO = 10 cm, BO = 6 cm and PB = 9 cm

As, OAQ ~OBP

⇒AQ = 15cm

Given: ACB ~ APQ

As, ACB and APQ are similar

Taking

⇒CA = 5.6cm

Now, taking

⇒AQ =3.25cm

Given: XY || BC

To find: XY

Since, XY || BC, AB is transversal, then,

AXY = ABC [by corresponding angles]

Since, XY || BC, AC is transversal, then,

AYX = ABC [by corresponding angles]

In AXY andABC

∠AXY = ∠ABC

∠AYX = ∠ACB

∴ AXY ≅ ABC [by AA similarity]

Since, triangles are similar, hence corresponding sides will be proportional

⇒XY = 1.5

Therefore, XY= 1.5cm

Given: ABC~PQR, PQ =20cm

And perimeter of ABC andPQR are 72cm and 48cm respectively.

As, ABC ~PQR

(corresponding sides are proportional)

⇒AB = 30cm

Given: PQ || RS

To Prove: POQ ~SOR

Let us takePOQ andSOR

∠OPQ = ∠OSR (as PQ || RS, Alternate angles)

∠POQ = ∠ROS (vertically opposite angles)

∠OQP = ∠ORS (as PQ || RS, Alternate angles)

∴ POQ ~SOR (by AAA similarity criterion)

Hence Proved

Given: ∠A = ∠C

To Prove: AOB ~COD

Let us take AOB andCOD

∠A = ∠C (given)

∠AOB = ∠COD (vertically opposite angles)

∴ AOB ~COD (by AA similarity criterion)

Hence Proved

We have, DB BC and AC BC

∠B + ∠C = 90° + 90°

⇒ ∠B + ∠C = 180°

∴ BD || AC

⇒∠EBD = ∠CAB (alternate angles)

Let us take BDE andABC

∠BED = ∠ACB (each 90°)

∠EBD = ∠CAB (alternate angles)

∴ BDE ~ABC (by AA similarity criterion)

Hence Proved

We have,

(∵, BD = DC as ∠1 = ∠2) …(i)

Also, ∠1 = ∠2

i.e. ∠DBC = ∠ACB

∴ ACB ~ DCE (by SAS similarity criterion)

Hence Proved

Given ABC is an isosceles triangle and AC = BC

∵ AC = BC

⇒∠CAB = ∠CBA

⇒180° – ∠CAB = 180° – ∠CBA

⇒ ∠CAP = ∠CBQ

Also, AP x BQ = AC²

(∵ AC =BC)

Thus, by SAS similarity, we get

ACP ~ BQC

Hence Proved

From the figure,

Hence,

Now it can be seen that both the triangles are similar as the corresponding sides are propotional.

From the figure we can see that,

∠ P = ∠ C

From ΔABC,

∠ A + ∠ B + ∠ C = 180°

60° + 80° + ∠ C = 180°

∠ C = 180° – 140°

∠ C = 40°

Hence, ∠ P = 40°

Here,

and

∴ PQ || BC [by converse of basic proportionality theorem]

Now, take APQ and ABC

∠APQ = ∠ABC (corresponding angles)

∠AQP = ∠ACB (corresponding angles)

∴ APQ ~ ABC (by AA similarity criterion)

Since, triangles are similar, hence corresponding sides will be proportional

⇒BC = 3PQ

Hence Proved

Here,

and

∴ PQ || BC [by the converse of basic proportionality theorem]

Now, take APQ and ABC

∠APQ = ∠ABC (corresponding angles)

∠AQP = ∠ACB (corresponding angles)

∴ APQ ~ ABC (by AA similarity criterion)

Since, triangles are similar, hence corresponding sides will be proportional

⇒BC = 4PQ

Hence Proved

In AOB and COD,

∠ AOB = ∠COD (Vertically opposite angles)

(given)

Therefore according to SAS similarity criterion,

∴ AOB ~ COD

Since, triangles are similar, hence corresponding sides will be proportional

⇒ DC = 10cm

Given: OA × OB = OC × OD

To Prove: A = C and B = D

Now, OA .OB = OC.OD

…(i)

In △AOD and △COB

(from (i))

∠AOD = ∠COB (vertically opposite angles)

∴ △AOD ~ △COB (by SAS similarity criterion)

We know that if two triangles are similar then their corresponding angles are equal.

⇒ A = C and B = D

Hence Proved

Given: CM is the median of △ABC and RN is the median of △PQR

Also, ABC ~ PQR

To Prove: (i) AMC ~PNR

CM is median of ABC

So, …(1)

Similarly, RN is the median of PQR

So, …(2)

Given ABC ~PQR

(corresponding sides of similar triangle are proportional)

(from (1) and (2))

…(3)

Also, since ABC ~PQR

∠A = ∠P …(4)

(corresponding angles of similar triangles are equal)

In AMC and PNR

∠A = ∠P (from (4))

(from (3))

∴ AMC ~PNR (by SAS similarity)

Hence Proved

(ii)To Prove:

In part (i), we proved that AMC ~PNR

So,

(corresponding sides of a similar triangle are proportional)

Therefore,

Hence Proved

(iii) CMB ~RNQ

Given ABC ~PQR

(corresponding sides of similar triangle are proportional)

(from (1) and (2))

…(5)

Also, since ABC ~PQR

∠B = ∠Q …(6)

(corresponding angles of similar triangles are equal)

In CMB and RNQ

∠B = ∠Q (from (6))

(from (5))

∴ CMB ~RNQ (by SAS similarity)

Hence Proved

Given: ABCD is a parallelogram

To Prove: DF x FE = BF x FA

In AFD and BFE

∠1 = ∠2 (alternate angles)

∠3 = ∠4 (vertically opposite angles)

∴ AFD ~BFE (by AA similarity criterion)

So,

(corresponding sides of similar triangle are proportional)

⇒ DF x FE = BF x FA

Hence Proved

(i) In ΔDAB and ΔACB

∠ADB = ∠CAB [each 90°]

∠DAB = ∠CAB [common angle]

∴ DAB ~ACB [by AA similarity]

Since the triangles are similar, hence corresponding sides are in proportional.

⇒ AB²= BC×BD

(ii) In ACB and DAC

∠CAB = ∠ADC [each 90°]

∠CAB = ∠CAD [common angle]

∴ ACB ~DAC [by AA similarity]

Since the triangles are similar, hence corresponding sides are in proportional.

⇒ AC² = BC. DC

(iii) In part (i) we proved that DAB ~ACB

⇒ AB × AC = BC × AD

Hence Proved

Given: ABC = 90° and BD AC

and AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm

To find: BC

Firstly, we have to show that ABC ~BDC

Let ABC and BDC

∠ABC = ∠BDC [each 90°]

∠ACB = ∠BCD [common angle]

∴ ABC ~BDC [by AA similarity criterion]

Since, triangles are similar, hence corresponding sides are proportional.

⇒ BC = 8.1cm

Given: CAB =90° and AD BC

and AC = 75 cm, AB = 1 cm and BC = 1.25 cm

Now, In ADB and CAB

∠ADB = ∠CAB [each 90°]

∠ABD = ∠CBA [common angle]

∴ ADB ~CAB [by AA similarity]

Since the triangles are similar, hence corresponding sides are in proportional.

⇒ AD = 60cm

Given: ΔABC ~ ΔDEF and AC = 3 cm and DF = 5 cm

To find: Areas of the two triangles

We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: AM = 6cm and DN = 9cm

Here, ΔABC and ΔDEF are similar triangles

We know that, in similar triangles, corresponding angles are in the same ratio.

⇒∠A = ∠D, ∠B = ∠E and ∠C = ∠F ……(i)

In ABM and DEN

∠B = ∠E [from (i)]

and ∠M = ∠N [each 90°]

∴ ABC ~ DEF [by AA similarity]

So, ……(ii)

We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

[from (ii)]

Given: ΔABC ~ ΔEF, BC = 3cm, EF = 4 cm

and area of ΔABC = 54 sq cm

We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

[given]

⇒ ar (∆DEF) = 96cm²

Given: ABC ~DEF, AB = 10cm,

and area of ABC = 20 sq cm , area of DEF = 45 sq cm

We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

[given]

⇒(DE)² = 5 × 45

⇒ DE = 15cm

Given: ΔABC ~ ΔADE

DE = 3cm, BC = 6 cm and area (ΔADE) =15 sq. cm

We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

[given]

⇒ ar (∆ABC) = 60cm²

Given: DE || BC

DE = 4cm, BC = 8cm and area (ADE) =25 sq. cm

In ABC andADE

∠B = ∠D [∵ DE || BC and AB is transversal,

Corresponding angles are equal]

∠C = ∠E [∵ DE || BC and AC is transversal,

Corresponding angles are equal]

∠BAC =∠DAE [common angle]

∴ ABC ~ADE [by AAA similarity]

Now, we know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

[given]

⇒ ar(ABC) = 25×4

⇒ ar (ABC) = 100cm²

Let ABC and DEF are two isosceles triangles with AB =AC and DE = DF and ∠A = ∠D

Now, let AM and DN are their respective altitudes or heights.

Let ABC and DEF

∠A = ∠D [given]

∴ ABC ~DEF [by SAS similarity]

We know that, in similar triangles, corresponding angles are in the same ratio.

⇒∠B = ∠E and ∠C = ∠F ……(i)

In ABM and DEN

∠B = ∠E [from (i)]

and ∠M = ∠N [each 90°]

∴ ABC ~ DEF [by AA similarity]

So, ……(ii)

We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

[from (ii)]

Given: Let ΔABC = 100cm² and ΔDEF = 49cm²

Let AM = 5cm

Here, ΔABC and ΔDEF are similar triangles

We know that, in similar triangles, corresponding angles are in the same ratio.

⇒∠B = ∠E and ∠C = ∠F …(i)

In ABM and DEN

∠B = ∠E [from (i)]

and ∠M = ∠N [each 90°]

∴ ABC ~ DEF [by AA similarity]

So, ……(ii)

We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

[from (ii)]

⇒DN = 3.5cm

The height of the other altitude is 3.5cm

Let ABC and DEF are two similar triangles such that ar (ABC) =100cm² and ar (DEF) = 64cm²

Also, let AM and DN are medians of ABC and DEF respectively.

Now in ABC and DEF

∠B = ∠E [∵ABC ~ DEF]

and

∴ ABC ~ DEF [by SAS similarity]

…(i)

Now, as ABC ~ DEF

We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

[from (i)]

⇒AM = 7cm

Hence, the length of the other median is 7cm.

Given: DE || BC

DE = 5cm, BC = 10cm and area (ADE) =20 sq. cm

In ABC andADE

∠B = ∠D [∵ DE || BC and AB is transversal,

Corresponding angles are equal]

∠C = ∠E [∵ DE || BC and AB is transversal,

Corresponding angles are equal]

∠BAC =∠DAE [common angle]

∴ ABC ~ADE [by AAA similarity]

Now, we know that the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

[given]

⇒ ar(ABC) = 20×4

⇒ ar (ABC) = 80cm²

Given: Let ABC = 81cm² and DEF = 49cm²

Let AM = 6.3cm

Here, ABC and DEF are similar triangles

We know that, in similar triangles, corresponding angles are in the same ratio.

⇒∠B = ∠E and ∠C = ∠F …(i)

In ABM and DEN

∠B = ∠E [from (i)]

and ∠M = ∠N [each 90°]

∴ ABC ~ DEF [by AA similarity]

So, …(ii)

We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

[from (ii)]

⇒DN = 4.9cm

Height of the other altitude is 4.9cm

Given: ΔABC ~ ΔDEF and AB =2DE

And area of ABC is 56 sq. cm

We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

[given]

⇒ ar(∆DEF) = 14sq cm

It is given that DE || BC and DE : BC = 4 : 5

Let ∆ ADE and ∆ABC

∠ADE = ∠ABC [corresponding angles]

∠AED = ∠ACB [corresponding angles]

∴ ∆ ADE ~ ∆ABC [by AA similarity]

We know that the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

Subtracting 1 from both the sides, we get

[given]

⇒

Given: AP= 1 cm, 1 BP= 3 cm, AQ = 1.5 cm, CQ = 4.5 cm

Here, and

⇒PQ || BC [by converse of basic proportionality theorem]

In ABC andAPQ

∠B = ∠P [∵ PQ || BC and AB is transversal,

Corresponding angles are equal]

∠C = ∠Q [∵ PQ || BC and AC is transversal,

Corresponding angles are equal]

∠BAC =∠PAQ [common angle]

∴ ABC ~APQ [by AAA similarity]

Now, we know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

[given]

Hence Proved

Given: AD ⊥ BC

and BC = 13 cm and AC = 5 cm

Let ABC and ADC

∠A = ∠D [each 90°]

∠C = ∠C [common angle]

∴ ABC ~ ADC [by AA similarity]

We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

(i) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.

Here, (8)² + (15)² = 64 + 225 = 289 = (17)²

∴ given sides 8cm, 15cm and 17cm make a right angled triangle.

(ii) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.

Here, (2a – 1)² + (2√(2a))²

⇒ 4a² + 1 – 4a + 8a

⇒ 4a² + 1 + 4a

= (2a + 1)²

∴ given sides (2a —1) cm, 2 cm and (2a + 1) cm make a right angled triangle.

(iii) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.

Here, (7)² + (24)² = 49 + 576 = 625 = (25)²

∴ given sides 7cm, 24cm and 25cm make a right angled triangle.

(iv) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.

Here, (1.4)² + (4.8)² = 1.96 + 23.04 = 25 = (5)²

∴ given sides 1.4cm, 4.8cm and 5cm make a right angled triangle.

Let AC be the position of a window from the ground and BC be the ladder, then the height of the window, AC =24m and length of the ladder, BC = 26m

Let AB = x m be the distance of the foot of the ladder from the base of the wall.

In ∆CAB, using Pythagoras Theorm,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AC)² + (AB)² = (BC)²

⇒ (24)² + (AB)² = (26)²

⇒ (AB)² = (26)² – (24)²

⇒ (AB)² = (26 – 24)(26+24)

[∵ (a² – b²)=(a+b)(a – b)]

⇒ (AB)² = (2)(50)

⇒ (AB)² = 100

⇒ AB = ±10

⇒ AB = 10 [taking positive square root]

Hence, the distance of the foot of the ladder from base of the wall is 10m

Let AB = 15m and AC = 8m

In ∆CAB, using Pythagoras Theorm,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AC)² + (AB)² = (BC)²

⇒ (8)² + (15)² = (BC)²

⇒ (BC)² = 64 + 225

⇒ (BC)² = 289

⇒ BC = ±17

⇒ BC = 17 [taking positive square root]

Hence, the man is 17m far from the starting point.

Let AC be the top of the building from the ground and BC be the ladder, then the height of the building, AC = 8m and length of the ladder, BC = 10m

Let AB = x m be the distance of the foot of the ladder from the building.

In ∆CAB, using Pythagoras Theorm,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AC)² + (AB)² = (BC)²

⇒ (8)² + (AB)² = (10)²

⇒ (AB)² = (10)² – (8)²

⇒ (AB)² = (10 – 8)(10+8)

[∵ (a² – b²)=(a+b)(a – b)]

⇒ (AB)² = (2)(18)

⇒ (AB)² = 36

⇒ AB = ±6

⇒ AB = 6 [taking positive square root]

Hence, the distance of the foot of the ladder from building is 6m

Let ABCD be a rectangle and AB and BC are the adjacent sides of length 30cm and 16cm respectively.

Let AC be the diagonal.

In ∆CBA, using Pythagoras Theorm,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AB)² + (BC)² = (AC)²

⇒ (30)² + (16)² = (AC)²

⇒ (AC)² = 900 + 256

⇒ (AC)² = 1156

⇒ AB = ±34

⇒ AB = 34 [taking positive square root]

Hence, the length of a diagonal of a rectangle is 34cm

Let AC be the position of a window from the ground and BC be the ladder, then the height of the window, AC =12m and length of the ladder, BC = 13m

Let AB = x m be the distance of the foot of the ladder from the base of the wall.

In ∆CAB, using Pythagoras Theorem,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AC)² + (AB)² = (BC)²

⇒ (12)² + (AB)² = (13)²

⇒ (AB)² = (13)² – (12)²

⇒ (AB)² = (13 – 12)(13+12)

[∵ (a² – b²)=(a+b)(a – b)]

⇒ (AB)² = (1)(25)

⇒ (AB)² = 25

⇒ AB = ±5

⇒ AB = 5 [taking positive square root]

Hence, the distance of the foot of the ladder from base of the wall is 5m

Let BC and AD be the two poles of height 14m and 9m respectively. Again, let CD be the distance between tops of the poles.

Then, CE = BC – AD = 14 – 9 = 5m [∵AD =BE]

Also, AB =12m

In ∆CED, using Pythagoras theorem, we get

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (CE)² + (DE)² = (CD)²

⇒ (5)² + (12)² = (CD)²

⇒ (CD)² = 25 + 144

⇒ (CD)² = 169

⇒ CD = √169

⇒ CD = ±13

⇒ CD = 13 [taking positive square root]

Hence, the distance between the tops of the poles is 13m

Let AB = 10m and AC = 24m

In ∆CAB, using Pythagoras Theorem,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AC)² + (AB)² = (BC)²

⇒ (24)² + (10)² = (BC)²

⇒ (BC)² = 576 + 100

⇒ (BC)² = 676

⇒ BC = ±26

⇒ BC = 26 [taking positive square root]

Hence, the man is 26m far from the starting point.

Let AB = 80m and AC = 150m

In ∆CAB, using Pythagoras Theorem,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AC)² + (AB)² = (BC)²

⇒ (150)² + (80)² = (BC)²

⇒ (BC)² = 22500 + 6400

⇒ (BC)² = 28900

⇒ BC = ±170

⇒ BC = 170 [taking positive square root]

Hence, the man is 170 m far from the starting point.

Given an isosceles triangle ABC with AC = BC, and AB² = 2AC²

To Prove: ∆ABC is a right triangle

Proof: AB² = 2AC² (given)

⇒ AB² = AC² +AC²

⇒ AB² = AC² +BC² [∵AC =BC]

⇒ ∆ABC is a right triangle right angled at C.

Let ABCD be a rhombus where AC = 10cm and BD =24cm

Let AC and BD intersect each other at O.

Now, we know that diagonals of rhombus bisect each other at 90°

Thus, we have

Since, AOB is a right angled triangle

So, by Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AO)² + (BO)² = (AB)²

⇒ (5)² + (12)² = (AB)²

⇒ (AB)² = 25 + 144

⇒ (AB)² = 169

⇒ AB = √169

⇒ AB = ±13

⇒ AB = 13 [taking positive square root]

Hence, AB = 13cm

Thus, length of each side of rhombus is 13cm

Given: ABC is an isosceles triangle right angled at C.

Let AC = BC

In ∆ACB, using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AC)² + (BC)² = (AB)²

⇒ (AC)² + (AC)² = (AB)²

[∵ABC is an isosceles triangle, AC =BC]

⇒ 2(AC)² = (AB)²

Hence Proved

Given: ABC is an isosceles triangle with AB = AC = 13 cm

Suppose the altitude from A on Bc meets BC at M.

∴ M is the midpoint of BC. AM = 5 cm

In ∆AMB, using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AM)² + (BM)² = (AB)²

⇒ (5)² + (BM)² = (13)²

⇒ (BM)² = (13)² – (5)²

⇒ (BM)² = (13 – 5)(13+5)

[∵ (a² – b²) = (a + b)(a – b)]

⇒ (BM)² = (8)(18)

⇒ (BM)² = 144

⇒ BM = ±12

⇒ BM = 12 [taking positive square root]

∴ BC = 2BM or 2MC = 2×12 = 24cm

Given: ABC is an equilateral triangle

∴ AB = AC = BC

and AD ⊥ BC

Now, In ∆ADB, using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AD)² + (BD)² = (AB)²

⇒ (AD)² + (BD)² = (BC)² [∵ AB = BC]

⇒ (AD)² + (BD)² = (2BD)² [ as AD⊥BC]

⇒ (AD)² + (BD)² = 4BD²

⇒ AD² = 3BD²

Given: ABC is an isosceles triangle

∴ AB = AC = 2a and BC = a

and AD is the altitude on BC. Therefore, BC = 2BD

Now, In ∆ADB, using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AD)² + (BD)² = (AB)²

[taking positive square root]

Given: ABC is an equilateral triangle

∴ AB = AC = BC = 2a

And let AD is an altitude on BC. Therefore,

Now, In ∆ADB, using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AD)² + (BD)² = (AB)²

⇒ (AD)² + (a)² = (2a)²

⇒ (AD)² = 4a² – a²

⇒ (AD)² = 3a²

⇒ AD = a√3 units

Given: ABC is an equilateral triangle

∴ AB = AC = BC = 12cm

And let AD is an altitude on BC. Therefore,

Now, In ∆ADB, using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AD)² + (BD)² = (AB)²

⇒ (AD)² + (6)² = (12)²

⇒ (AD)² = 144 – 36

⇒ (AD)² = 108

⇒ AD = √108

⇒ AD = 6√3

Hence, the height of an equilateral triangle is 6√3 cm

Given: ABC is a right triangle right angled at B

and L and M are the mid-points of AB and BC respectively.

⇒ AL = LB and BM = MC

In ∆LBC, using Pythagoras theorem we have,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (LB)² + (BC)² = (LC)²

⇒ (AB)² + 4(BC)² = 4(LC)²

Hence Proved

Let ABCD be a rhombus having AD = 5cm and AC = 6cm

Now, we know that diagonals of rhombus bisect each other at 90°

Thus, we have

Since, AOD is a right angled triangle

So, by Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AO)² + (BO)² = (AD)²

⇒ (3)² + (BO)² = (5)²

⇒ (BO)² = 25 – 9

⇒ (BO)² = 16

⇒ BO = √16

⇒ BO = ±4

⇒ BO = 4 [taking positive square root]

Hence, BO = 4cm

⇒ BC = 2BO = 2 × 4 = 8cm

Thus, length of each side of rhombus is 13cm.

Given: B = 90° and D is the midpoint of BC .i.e. BD = DC

To Prove: AC² = AD² + 3CD²

In ∆ABC, using Pythagoras theorem we have,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AB)² + (BC)² = (AC)²

⇒ (AB)² + (2CD)² =(AC)²

⇒ (AB)² + 4(CD)² =(AC)²

⇒ (AD² – BD²) + 4(CD²) = AC²

[∵ In right triangle ∆ABD, AD² =AB² + BD² ]

⇒ AD² – BD² + 4CD² = AC²

⇒ AD² – CD² + 4CD² = AC²

[∵ D is the midpoint of BC, BD = DC]

⇒ AD² +3CD² = AC²

or AC² = AD² + 3CD²

Hence Proved

Given: C = 90° and D is the midpoint of BC .i.e. BC = 2CD

To Prove: AB² = 4AD² — 3AC²

In ∆ABC, using Pythagoras theorem we have,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AC)² + (BC)² = (AB)²

⇒ (AC)² + (2CD)² =(AB)²

⇒ (AC)² + 4(CD)² =(AB)²

⇒ (AC)² + 4(AD² – AC²) = AB²

[∵ In right triangle ∆ACD, AD² =AC² + CD² ]

⇒ AC² +4AD² – 4AC² = AB²

⇒ 4AD² – 3AC² = AB²

or AB² = 4AD² — 3AC²

Hence Proved

Given: AB = AC and BD AC

To Prove: BD² – CD² = 2CD × AD

In ∆BDC, using Pythagoras theorem we have,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (BD)² + (CD)² = (BC)² …(i)

In ∆BDA, using Pythagoras theorem we have,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (BD)² + (AD)² = (AB)²

⇒ (BD)² + (AD)² = (AC)² [∵ AB = AC]

Multiply this eq. by 2, we get

⇒ 2(BD)² + 2(AD)² = 2(AC)² …(ii)

Subtracting Eq. (ii) from (i), we get

⇒ CD² – BD² = BC² – 2 AC² + 2 AD²

= BC² – 2 (AD +CD)² + 2 AD²

= BC² – 2 CD² – 4 AD × CD

= BD² + CD² – 2 CD² – 4 AD × CD

= BD² – CD² – 4 AD × CD

⇒ CD² – BD² –BD² +CD² = –4AD × CD

⇒ –2(BD² – CD²) = –4AD × CD

⇒ BD² – CD² = 2CD × AD

Hence Proved

Given: ABCD is a quadrilateral and B = 90°

and AD²= AB² + BC² + CD²

To Prove: ∠ACD = 90°

In right triangle ∆ABC, using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AB)² + (BC)² = (AC)² …(i)

Given: AD²= AB² + BC² + CD²

⇒ AD²= AC² + CD² [from (i)]

In ∆ACD

AD²= AC² + CD²

∴ ∠ACD = 90° [converse of Pythagoras theorem]

Hence Proved

In rhombus ABCD, AB = BC = CD = DA

We know that diagonals bisect each other at 90°

And

Consider right triangle AOB

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (OA)² + (OB)² = (AB)²

⇒ AC² + BD² = 4AB²

⇒ AC² + BD² = AB² + AB² + AB² + AB²

⇒ AC² + BD² = AB² + BC² + CD² + DA²

Hence Proved

Given: ABC is an equilateral triangle

and AD is the altitude on side BC

To Prove: 3AB²= 4AD²

In right triangle ∆ADB, using Pythagoras theorem

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ AD² + BD² = AB²

⇒ 4AD² + BC² = 4AB²

⇒ 4AD² = 4AB² – BC²

⇒ 4AD² = 4AB² – AB² [∵ABC is an equilateral triangle]

⇒ 4AD² = 3AB²

Hence Proved

Construction: Draw an altitude from A on BC and named it O.

Given: ABC is an isosceles triangle with AB = AC

To Prove: AD² —AC² = BD × CD

In right triangle ∆AOD, using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ AO² + OD² = AD² …(i)

Now, in right triangle ∆AOB, using Pythagoras theorem, we have

⇒ AO² + BO² = AB² …(ii)

Subtracting eq (ii) from (i), we get

AD² – AB² = AO² + OD² – AO² – BO²

⇒ AD² – AB² = OD² – BO²

⇒ AD² – AB² = (OD + BO)(OD – OB)

[∵ (a² – b²)= (a + b)(a – b)]

⇒ AD² – AB² = (BD)(OD – OC) [∵OB = OC]

⇒ AD² – AB² = (BD)(CD)

⇒ AD² – AC² = (BD)(CD) [∵AB =AC]

Hence Proved

Given: In ABC, D is the mid-point of BC and AE BC

and AC > AB

In right triangle ∆AEB, using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AE)² + (BE)² = (AB)²

⇒ (AE)² + (BD – ED)² = (AB)²

⇒ (AE)² + (ED)² + (BD)² – 2 (ED)(BD) = (AB)²

[∵ (a – b)² = a² + b² – 2ab]

⇒ (AE² + ED²) + (BD)² – 2 (ED)(BD) = (AB)²

⇒ (AD)² + (BD)² – 2 (ED)(BD) = (AB)²

[∵ In right angled ∆AED, AE² + ED² =AD²]

[∵ D is the midpoint of BC, so 2DC = BC]

Hence Proved

Given ∆ABC is an isosceles triangle in which ∠B is right angled i.e. 90°

⇒ AB = BC

In right angled ∆ABC, by Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AB)² + (BC)² = (AC)²

⇒ (AB)² + (AB)² = (AC)²

[∵ABC is an isosceles triangle, AB =BC]

⇒ 2(AB)² = (AC)²

⇒ (AC)² = 2(AB)² …(i)

It is also given that ∆ABE ~ ∆ADC

And we also know that, the ratio of similar triangles is equal to the ratio of their corresponding sides.

[from (i)]

∴ ar(∆ABE) : ar(∆ADC) = 1 : 2

Given: ∠POR = 90°, OP = 6 cm and OR= 8 cm

and PQ = 24 cm and QR = 26 cm

To Prove: PQR is right angled at P

In ∆POR, using Pythagoras theorem, we get

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (PO)² + (OR)² = (PR)²

⇒ (6)² + (8)² = (PR)²

⇒ 36 +64 = (PR)²

⇒ (PR)²= 100

⇒ PR =√100

⇒ PR = 10 [taking positive square root]

In ∆PQR

Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.

Here, (PR)² + (PQ)²

⇒ (10)² + (24)²

= 100 + 576

= 676

= (26)² = (QR)²

∴ given sides 10cm, 24cm and 26cm make a right triangle right angled at P.

Hence Proved

Given: D is the mid-point of side BC and AE BC

and BC = a, AC = b, AB= c, ED = x, AD =p and AE = h

To Prove: (i)

or

Proof: In right triangle ∆AEC, using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AE)² + (EC)² = (AC)²

⇒ (AE)² + (ED + DC)² = (AC)²

⇒ (AE)² + (ED)² + (DC)² + 2 (ED)(DC) = (AC)²

[∵ (a + b)² = a² + b² +2ab]

⇒ (AE² + ED²) + (DC)² + 2 (ED)(DC) = (AC)²

⇒ (AD)² + (DC)² + 2 (ED)(DC) = (AC)²

[∵ In right angled ∆AED, AE² + ED² =AD²]

[∵ D is the midpoint of BC, so 2DC = BC]

…(i)

To Prove: (ii)

or

Proof: In right triangle ∆AEB, using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AE)² + (BE)² = (AB)²

⇒ (AE)² + (BD – ED)² = (AB)²

⇒ (AE)² + (ED)² + (BD)² – 2 (ED)(BD) = (AB)²

[∵ (a – b)² = a² + b² – 2ab]

⇒ (AE² + ED²) + (BD)² – 2 (ED)(BD) = (AB)²

⇒ (AD)² + (BD)² – 2 (ED)(BD) = (AB)²

[∵ In right angled ∆AED, AE² + ED² =AD²]

[∵ D is the midpoint of BC, so 2DC = BC]

…(ii)

On adding eq. (i) and (ii), we get

To Prove: (iii) (b² —c²) = 2ax

or (AC)² – (AB)² = 2 (BC)(ED)

Proof: Subtracting Eq. (ii) from (i), we get

⇒ (AC)² – (AB)² = 2 (BC)(ED)

Hence Proved

Given: ABC ia right triangle right angled at C

P and Q are the mid–points of the sides CA and CB respectively.

⇒ AP = PC and CQ = QB

In ACB, using Pythagoras Theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AC)² + (BC)² = (AB)² …(i)

Now, In ACQ, using Pythagoras Theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AC)² + (CQ)² = (AQ)²

⇒ 4(AC)² + (BC)² = 4(AQ)²

⇒ (BC)² = 4(AQ)² – 4(AC)² …(ii)

Now, In PCB, using Pythagoras Theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (PC)² + (BC)² = (BP)²

⇒ (AC)² + 4(BC)² = 4(BP)²

⇒ (AC)² = 4(BP)² – 4(BC)² …(ii)

Putting the value of (AC)² and (BC)² in eq. (i), we get

4(BP)² – 4(BC)² + 4(AQ)² – 4(AC)² = (AB)²

⇒ 4(BP² +AQ²) – 4(BC² + AC²) = (AB)²

⇒ 4(BP² +AQ²) – 4(AB²) = (AB)² [from eq(i)]

⇒ 4(BP² +AQ²) = 5(AB)²

Hence Proved