Equal Triangles

To find all pairs of matching angles, we always look for the corresponding sides or similar sides in the two triangles and the angles between any two of the corresponding sides in the two figures are matching angles.

In the triangles given above, we first find the corresponding sides or similar sides.

AC = RQ = 4 cm

CB = PQ = 5 cm

AB = PR = 6 cm

To find pairs of matching angles, the angles between any two of the corresponding sides in the two triangles are matching angles.

a) Angle between AC(4 cm) and AB(6 cm), ∠CAB = Angle between RQ(4 cm) and PR(6 cm) , ∠PRQ.

b) Angle between AC(4 cm) and CB(5 cm), ∠ACB = Angle between RQ(4 cm) and PQ(5 cm) , ∠RQP.

c) Angle between CB(5 cm) and AB(6 cm), ∠CBA = Angle between PQ(5 cm) and PR(6 cm) , ∠QPR.

To find all pairs of matching angles, we always look for the corresponding sides or similar sides in the two triangles and the angles between any two of the corresponding sides in the two figures are matching angles.

In the triangles given above, we first find the corresponding sides or similar sides.

NL = XY = 4 cm

LM = YZ = 8 cm

NM = XZ = 10 cm

To find pairs of matching angles, the angles between any two of the corresponding sides in the two triangles are matching angles.

a) Angle between NL (4 cm) and LM (8 cm), ∠NLM = Angle between XY (4 cm) and YZ (8 cm) , ∠XYZ.

b) Angle between LM (8 cm) and NM (10 cm), ∠LMN = Angle between YZ (8 cm) and XZ (10 cm) , ∠YZX.

c) Angle between NM (10 cm) and NL (4 cm), ∠MNL = Angle between XZ (10 cm) and XY (4 cm) , ∠ZXY.

We know sum of all interior angles of a triangle is 180⁰.

In ∆ABC,

∠a + ∠b + ∠c = 180⁰

40⁰ + 60⁰ + ∠c = 180⁰

100⁰ + ∠c = 180⁰

∠c = 180⁰-100⁰

∠c = 80⁰

To find all angles in ∆PQR:

Given, AB = QR

BC = RP

CA = PQ

To find pairs of matching angles, the angles between any two of the corresponding sides in the two figures are matching angles.

Thus,

∠CAB = ∠PQR = 40⁰

∠ABC = ∠QRP = 60⁰

∠BCA = ∠RPQ = 80⁰

Given, AB = QR

BC = PQ

CA = RP, which are corresponding sides of both triangles.

Thus, angles between any two of the corresponding sides in the two figures are matching angles.

Finding matching angles will help us to compute the remaining angles.

Therefore,

∠ABC = ∠PQR = 70⁰

∠PRQ = ∠CAB = 60⁰

Now, we have,

In ∆ABC,

∠a + ∠b + ∠c = 180⁰

60⁰ + 70⁰ + ∠c = 180⁰

130⁰ + ∠c = 180⁰

∠c = 180⁰-130⁰

∠c = 50⁰

In ∆PQR,

∠p + ∠q + ∠r = 180⁰

∠p + 70⁰ + 60⁰ = 180

∠p + 130⁰ = 180⁰

∠p = 180⁰-130⁰

∠p = 50⁰

Since AC = AD and CB = DB, ∠ACB = ∠ADB.

As both the triangles have AB as a common side, angle between any side and AB in ∆ABC will be a matching angle to angle between

any side and AB in ∆ABD.

So, AC = AD and AB which is common, ∠CAB = ∠CAB.

And, CB = DB and AB which is common, ∠CBA = ∠DBA.

Hence, all the angles of ∆ABC and ∆ABD are equal.

Given, AB = AD

BC = CD

AC which is common to both ∆ADC and ∆ABC.

So, ∠DAC = ∠CAB = 30 as AB = AD and AC which is common.

And, ∠DCA = ∠ACB = 50 as BC = CD and AC which is common.

Now, we know sum of all interior angles of a triangle is 180⁰.

In ∆ADC,

∠a + ∠d + ∠c = 180⁰

30⁰ + ∠d + 50⁰ = 180

80⁰ + ∠d = 180⁰

∠d = 180⁰-80⁰

∠d = 100⁰

In ∆ABC,

∠a + ∠b + ∠c = 180⁰

30⁰ + ∠b + 50⁰ = 180⁰

80⁰ + ∠b = 180⁰

∠b = 180⁰-80⁰

∠b = 100⁰

To find all pairs of matching angles, we always look for the corresponding sides or similar sides in the two triangles and the angles between any two of the corresponding sides in the two figures are matching angles.

(i)

We can see that ∠CAB = ∠PQR = 40⁰ as shown in the figure.

Angle between AC (7 cm) and CB, ∠ACB = Angle between PQ(7 cm) and PR , ∠QPR.

Angle between AB (3 cm) and CB, ∠ABC = Angle between RQ(3 cm) and PR , ∠PRQ.

(ii)

We can see that ∠NLM = ∠XZY = 50⁰ as shown in the figure.

Angle between LN (4 cm) and NM, ∠LNM = Angle between XZ (4 cm) and XY, ∠ZXY

Angle between LM (6 cm) and NM, ∠ LMN = Angle between YZ(6 cm)and XY , ∠ZYX

AC = BE = 6 cm

AB = BD = 4 cm

(i) Yes, the length of BC and DE is equal because they are the corresponding sides of two equal triangles.

(ii) Yes, BC and DE are parallel because they are equal and corresponding sides of two equal triangles.

We can see that the diagonal AB makes the same angle of 35⁰ with each AC(6 cm) and DB(6 cm). Hence, it acts as a transversal

between AC and DB which makes AC and DB parallel and equal lines. Hence, ABCD is a parallelogram.

Given, M is the midpoint of the line AB and ∠m = 90⁰.

∠a = 50

We know sum of all interior angles of a triangle is 180⁰.

In ∆AMC,

∠a + ∠b + ∠c = 180⁰

50⁰ + 90 + ∠c = 180⁰

140⁰ + ∠c = 180⁰

∠c = 180⁰-140⁰

∠c = 40⁰

In ∆MBC,

∠c = 40⁰

∠m = 90⁰

∠b = ?

∠m + ∠b + ∠c = 180⁰

90⁰ + ∠b + 40⁰ = 180⁰

130⁰ + ∠b = 180⁰

∠b = 180⁰-130⁰

∠b = 50⁰

∠a = 50⁰, ∠b = 50⁰, ∠c = 80⁰

Given, lines AB and CD are parallel and M is the midpoint of AB.

As AB and CD are parallel, AD, DM, MC and CB are transversal.

(i) ∠AMD = ∠MBC = 60⁰ (corresponding angles)

∠CMB = ∠DAM = 40⁰ (corresponding angles)

∠CDM = AMD = 60⁰ (alternate interior angles)

∠DCM = ∠CMB = 40⁰ (alternate interior angles)

On straight line AMB,

∠ AMD + ∠DMC + ∠CMB = 180⁰ (angles in a straight line)

60⁰ + ∠DMC + 40⁰ = 180⁰

∠DMC + 100⁰ = 180⁰

∠DMC = 180⁰-100⁰

∠DMC = 80⁰

∠ADM = ∠DMC = 80⁰ (alternate interior angles)

∠MCB = ∠DMC = 80⁰ (alternate interior angles)

Therefore, now we have,

(ii) Quadrilaterals AMCD and MBCD both contain two equal triangles. That is what makes special.

Quad. AMCD consists of ∆AMD and ∆DMC.

Quad. MBCD consists of ∆MBC and ∆DMC.

In Δ ABC and Δ PQR,

(AB) = l(RQ) = 6 cm

∠A = ∠R = 50°

∠B = ∠Q = 70°

According to property,

If one side of a triangle and angle at its ends are equal to one side of another triangle and the angles at its ends, then the third angles are also equal and the sides opposite equal angle are equal.

In Δ ABC and Δ PQR,

BC is opposite ∠A and PQ is opposite ∠R

Also, AC is opposite ∠B and PR is opposite ∠Q

∴ BC = PQ

And AC = PR

Also, AB = RQ = 6 cm (given) opposite \angle l anglee

In Δ LMN and Δ XYZ,

LN = XY = 8 cm

∠L = ∠Y = 30°

In Δ LMN, sum of all angles of triangle is 180°

∴ ∠L + ∠M + ∠N = 180°

30 + 80 + ∠N = 180

110 + ∠N = 180

∴ ∠N = 180-110

= 70°

Here, ∠N = ∠X = 70°

According to property,

If one side of a triangle and angle at its ends are equal to one side of another triangle and the angles at its ends, then the third angles are also equal and the sides opposite equal angle are equal.

{Here, LN and XY is the required side}

In Δ LMN and Δ XYZ,

NM is opposite ∠L and XZ is opposite ∠Y

Also, LM is opposite ∠N and YZ is opposite ∠X

∴ NM = XZ

And LM = YZ

Also, LN = XY = 8 cm (given)

Given-

AP = BQ

And AP ∥ BQ

According to the property,

If a pair of equal and opposite sides is parallel, then the four points connected form a parallelogram.

∴ the imaginary quadrilateral APBQ is a parallelogram.

AB and PQ are diagonals of parallelogram.

Also, the diagonals of parallelogram bisect each other.

∴ PM = MQ

And AM = MB

Also, ∠AMP = ∠BMQ, (vertically opposite angles)

According to property,

When two sides of a triangle and angle made by them are equal to the two sides and angle made by them of another triangle, then the third sides and the corresponding two angles are also equal.

∴ AP = BQ

Hence, the sides of the two triangles are equal.

ii) Specialty of point M,

It divides both sides in equal ratio i.e. 1:1

(Reason, Discussed above i.e. the imaginary quadrilateral APBQ is a parallelogram and AP,BQ are its diagonals).

iii) Steps for construction,

1. Draw a line AB = 5.5 cm

2. Using set-square draw two perpendicular lines on AB at each points A and B of length 5.5 cm as AP( = 5.5 cm) and BQ( = 5.5 cm)

3.Join lines PB and AQ and let them intersect at point O.

4.Drop a perpendicular from O to AB and let the new point be M.

M is the required midpoint of AB.

Here, all angles of the polygon are equal(a regular polygon).

∴ ∠ABC = ∠AED …(eq)1

(∠PBC,∠ABC) and (∠AED,∠QED) form a linear pair.

∠PBC = 180-∠ABC (Linear angles are supplementary)

= 180-∠AED (from eq1)

∠PBC = ∠QED …(eq)2 (∠AED and ∠QED form linear pair).

∠BCD = ∠EDC …(eq)2

(∠BCD, ∠BCP) and (∠EDC,∠EDQ) form a linear pair.

∠BCP = 180-∠BCD (Linear angles are supplementary)

= 180-∠EDC (from eq2)

∠BCP = ∠EDQ …(eq)3

According to property,

If one side of a triangle and angle at its ends are equal to one side of another triangle and the angles at its ends, then the third angles are also equal and the sides opposite equal angle are equal.

In ΔPBC and Δ QED,

∠BCD = ∠EDC

∠BCP = ∠EDQ

Also, BC = DE (a regular polygon)

Hence, PC = DQ (∠PBC = ∠DEQ)

PB = EQ (∠PCB = ∠EDQ)

And ∠BPC = ∠EQD …(eq)4

Hence, the sides are equal.

ii) From eq4,

∠BPC = ∠EQD

∴ ∠APQ = ∠AQP

Hence, AP = AQ (converse of isosceles angle theorem).

In Δ ABC and Δ PQR

AB = QR

BC = RP

CA = PQ

∴ Δ ABC ≅ Δ QRP {the corresponding sides on both sides of the equation are equal}

∴ ∠ABC = ∠QRP …(eq)1

In Δ CDB and Δ PSR,

∠DBC(∠ABC) = ∠SRP(∠QRP) …(eq)2 (from eq1)

∠CDB = ∠PSR = 90°

Thus, in these triangles 2 angles are equal

Since, sum of all angles of a triangle is 180°

∴ third angle is also equal

∴ ∠DCB = ∠RPS …(eq)3

According to property,

If one side of a triangle and angle at its ends are equal to one side of another triangle and the angles at its ends, then the third angles are also equal and the sides opposite equal angle are equal.

{Here, common side is BC and RP

∠DCB = ∠RPS (from eq3)

∠DBC = ∠SRP (from eq2)}

Applying it in Δ CDB and Δ PSR

CD = PS …(eq)4 (CD is opposite ∠CBD,PS is opposite ∠PRS and ∠CBD = ∠PRS)

ii) For a triangle,

∴ area(Δ ABC) =

And area (ΔPQR) =

Here, AB = QR (given)

And, CD = PS (from eq4)

Hence, areas of both the triangles are equal.

Given-

AB ∥ CD

BM = MC

In Δ DCM and Δ BMN,

∠DMC = ∠BMN (vertically opposite angles)

BM = MC

∠DCM = ∠MBN (alternate angle test, DC ∥ BN with BC as transversal).

Using the property,

If one side of a triangle and angle at its ends are equal to one side of another triangle and the angles at its ends, then the third angles are also equal and the sides opposite equal angle are equal.

∴ ∠CDM = ∠MNB

DM = MN

DC = BN

As all 3 sides and angles are equal.

Δ DCM ≅ Δ NBM

Also, congruent triangles have equal areas.

∴ area(Δ DCM) = area(BMN) …(eq)1

area of a parallelogram =

area of a triangle =

area of parallelogram = area(ABCD)

= area(ADMB) + area(Δ DCM) …(eq)1

Area of triangle = area(Δ AND)

= area(ADMB) + area(Δ BMN)

= area(ADMB) + area(Δ DCM) (from eq1) …(eq)2

∴ area of parallelogram = area of triangle

A rectangle is a parallelogram where each angle is a right angle.

Here, AD = BC …(eq)1

And AB = CD

According to Pythagoras theorem,

(Hypotenuse)² = (one side)² + (other side)²

In Δ ABD,

Hypotenuse = BD

One side = AB

And other side = AD

∴ (BD)² = (AB)² + (AD)² …(eq)2

In Δ ABC,

Hypotenuse = AC

One side = AB

And other side = BC

∴ (AC)² = (AB)² + (BC)²

= (AB)² + (AD)² …(eq)3 …(from eq1)

As is clear from (eq)2 and (eq)3

BD² = AC²

Taking square root on both sides,

BD = AC

∴ Diagonals of a rectangle are equal.

In first figure,

As seen it seems AB = BC (since it is a isosceles triangle).

Also, if 2 sides of a triangle are equal, the angles opposite equal sides are also equal.

∴ ∠BAC(∠A) = ∠BCA(∠C) = 30°

In a triangle,

Sum of all angles of a triangle is 180°

Hence, in Δ ABC,

∠A + ∠B + ∠C = 180°

∴ 30 + ∠B + 30 = 180°

∠B + 60 = 180°

∴ ∠B = 180-60

= 120°

In second figure,

As seen it seems DE = DF (since it is a isosceles triangle).

Also, if 2 sides of a triangle are equal, the angles opposite equal sides are also equal.

∴ ∠DEF(∠E) = ∠DFE(∠F) = y° …(eq)1

In a triangle,

Sum of all angles of a triangle is 180°

Hence, in Δ DEF,

∠D + ∠E + ∠F = 180°

∴ 40 + y + y = 180°

∴ 40 + 2y = 180

∴ 2y = 180-40

= 140°

∴

Hence, ∠E = ∠F = 70° (from eq1)

In 3^rd figure,

As seen it seems PQ = PR (since it is a isosceles triangle).

Also, if 2 sides of a triangle are equal, the angles opposite equal sides are also equal.

∴ ∠PQR(∠Q) = ∠PRQ(∠R) = y° …(eq)1

In a triangle,

Sum of all angles of a triangle is 180°

Hence, in Δ PQR,

∠P + ∠Q + ∠R = 180°

∴ 20 + y + y = 180°

∴ 20 + 2y = 180

∴ 2y = 180-20

= 160°

∴

Hence, ∠Q = ∠R = 80° (from eq1)

In 4^th figure,

As seen it seems XY = XZ (since it is a isosceles triangle).

Also, if 2 sides of a triangle are equal, the angles opposite equal sides are also equal.

∴ ∠XYZ(∠Y) = ∠XZY(∠Z) = m° …(eq)1

In a triangle,

Sum of all angles of a triangle is 180°

Hence, in Δ XYZ,

∠X + ∠Y + ∠Z = 180°

∴ 100 + m + m = 180°

∴ 100 + 2m = 180

∴ 2m = 180-100

= 80

∴

Hence, ∠Y = ∠Z = 40° (from eq1)

Let in Δ ABC,

∠A = 90°

It is given that triangle is a isosceles triangle.

Case 1

Two equal angles both equal to 90°

If it happens, then sum of two equal angles will be = 90 + 90 = 180°

But, sum of all angles of a triangle is 180°

Thus, the third angle = 180-180 = 0

But, this is not possible.

Case 2

Two equal angles other than right angle.

If ∠A = 90°

Let ∠B = ∠C = y°

Sum of all angles of a triangle is 180°

∴ ∠A + ∠B + ∠C = 180°

90 + y + y = 180°

90 + 2y = 180°

2y = 180-90

= 90

∴

Hence the other 2 angles are 45°

Let in Δ ABC,

∠A = 60°

It is given that triangle is a isosceles triangle.

Case 1

Two equal angles both equal to 60°

If it happens, then sum of two equal angles will be = 60 + 60 = 120°

But, sum of all angles of a triangle is 180°

Thus, the third angle = 180-120 = 60

Hence, other two angles are 60° each.

Case 2

Two equal angles other than given angle (∠A)

If ∠A = 60°

Let ∠B = ∠C = y°

Sum of all angles of a triangle is 180°

∴ ∠A + ∠B + ∠C = 180°

60 + y + y = 180°

60 + 2y = 180°

2y = 180-60

= 120

∴

Hence the other 2 angles are also 60°

Hence, in both the cases the triangle is equilateral (each angle = 60°)

In Δ OAB,

OA = OB = r = Radius of circle

In a triangle with 2 equal sides, the angles opposite to equal sides are equal.

∴ ∠OAB(∠A) = ∠OBA(∠B) = y° …(eq)1

In a triangle,

Sum of all angles of a triangle is 180°

∴ In Δ OAB,

∠O + ∠A + ∠B = 180°

∴ 60 + y + y = 180°

∴ 60 + 2y = 180°

∴ 2y = 180-60

= 120

∴

∴ ∠A = ∠B = 60° (from eq1)

Here,

∠AOB = ∠AOC = 120°

Sum of a complete whole angle is 360°

∴ ∠AOB + ∠AOC + ∠BOC = 360°

120 + 120 + ∠BOC = 360°

∴ 240 + ∠BOC = 360°

∴ ∠BOC = 360-240

= 120°

In Δ OAB,

OA = OB = radius of circle

In a triangle with 2 equal sides, the angles opposite to equal sides are equal.

∴ ∠OAB = ∠OBA = y°

In a triangle,

Sum of all angles of a triangle is 180°

Hence, in Δ OAB,

∠OAB + ∠OBA + ∠AOB = 180°

∴ y + y + 120 = 180°

∴ 120 + 2y = 180

∴ 2y = 180-120

= 60

∴

∴ ∠OAB = ∠OBA = 30° …(eq)1

In Δ OAC,

OA = OC = radius of circle

In a triangle with 2 equal sides, the angles opposite to equal sides are equal.

∴ ∠OAC = ∠OCA = y°

In a triangle,

Sum of all angles of a triangle is 180°

Hence, in Δ OAB,

∠OAC + ∠OCA + ∠AOC = 180°

∴ y + y + 120 = 180°

∴ 120 + 2y = 180

∴ 2y = 180-120

= 60

∴

∴ ∠OAC = ∠OCA = 30° …(eq)2

In Δ OBC,

OB = OC = radius of circle

In a triangle with 2 equal sides, the angles opposite to equal sides are equal.

∴ ∠OCB = ∠OBC = y°

In a triangle,

Sum of all angles of a triangle is 180°

Hence, in Δ OCB,

∠OCB + ∠OBC + ∠COB = 180°

∴ y + y + 120 = 180°

∴ 120 + 2y = 180

∴ 2y = 180-120

= 60

∴

∴ ∠OAC = ∠OBC = 30° …(eq)3

From equations 1,2 and 3

In Δ ABC

∠A = ∠BAO + ∠OAC

= 30 + 30

= 60°

∠B = ∠OBA + ∠OBC

= 30 + 30

= 60°

∠C = ∠OCB + ∠OCA

= 30 + 30

= 60°