Congruency Of Triangles

Corresponding Sides of the two triangles are

PQ = XY

QR = YZ

RP = ZX

Corresponding Angles of the two triangles are

∠PQR = ∠XYZ

∠PRQ = ∠XZY

∠QPR = ∠YXZ

Corresponding Sides of the two triangles are

PQ = AB

QR = BC

RP = CA

Corresponding Angles of the two triangles are

∠PQR = ∠ABC

∠PRQ = ∠ACB

∠QPR = ∠CAB

In the adjoining figure if ∠C = ∠F, then AB = DE and BC = EF.

Since the triangles are congruent the corresponding sides of the triangles are equal

Since ∠C = ∠F

AB = DE

BC = EF

In the adjoining figure if BC = EF, then ∠C = ∠F and ∠A = ∠D.

Since the triangles are congruent the corresponding angles of the triangles are equal

Since BC = EF

∠C = ∠F

∠A = ∠D

In the adjoining figure, if AC = CE and ΔABC ≅ ΔDEC, then ∠D = ∠B and ∠A = ∠E.

Since the triangles ΔABC and ΔDEC are congruent the corresponding angles of the triangles are equal

Since AC = CE

∠D = ∠B

∠A = ∠E

In Δ POQ and Δ SOR

OP = OR(Diagonals of a rectangle bisect each other)

OQ = OS(Diagonals of a rectangle bisect each other)

PQ = SR(Diagonals are equal in length)

Δ POQ ≅ Δ SOR by S.S.S. axiom of congruency

In Δ POS and Δ QOR

OP = OR(Diagonals of a rectangle bisect each other)

OQ = OS(Diagonals of a rectangle bisect each other)

PS = QR(Diagonals are equal in length)

Δ POS ≅ Δ QOR by S.S.S. axiom of congruency

In Δ PSQ, Δ PQR, Δ QRS and Δ PRS

PS = QR = QR = PS(Opposite sides of rectangle)

PQ = PQ = SR = SR(Opposite sides of the rectangle)

∠P = ∠Q = ∠R = ∠S (Angles of a rectangle)

Δ PSQ ≅ Δ PQR ≅ Δ QRS ≅ Δ PRS by S.A.S. axiom of congruency

In Δ APM, Δ BMN, Δ CNO and Δ DOP

AB = BC = CD = DA (Sides of a square)

Since M, N, O, and P are the midpoints of sides AB, BC, CD and DA

⇒ 2AM = 2BN = 2CO = 2DP

⇒ AM = BN = CO = DP …(i)

⇒ 2AP = 2BM = 2CN = 2DO

⇒ AP = BM = CN = DO …(ii)

∠PAM = ∠MBN = ∠NCO = ∠ODP(Angles of a square) …(iii)

So Δ APM ≅ Δ BMN ≅ Δ CNO ≅ Δ DOP by S.A.S. axiom of congruency.

In Δ BCD and Δ CBE we have

AB = AC

AE = AD

Subtracting the above two equations we get

AB-AE = AC-AD

⇒ Be = CD

∠EBC = ∠DCB(Base angle of an isosceles triangle)

BC = BC(Common side)

So Δ BCD and Δ CBE are congruent to each other by S.A.S. axiom of congruency.

In Δ AED and Δ ABC we have

BA = AD(Given)

CA = AE(Given)

∠EAD = ∠BAC (Vertically Opposite)

So Δ AED and Δ ABC are congruent by S.A.S. axiom of congruency

So we can say

∠AED = ∠ACB(Corresponding parts of Congruent triangles)

So ∠AED & ∠ACB forms a pair of alternate interior angles

Hence DE || BC

In Δ ABC since AB = AC , so the triangle is isosceles

In an isosceles triangle, the base angles are always equal

So ∠B = ∠C = x(Let’s assume)

Since the sum of interior angles of a triangle is 180⁰

∠A + ∠B + ∠C = 180⁰

⇒ 2x = 180⁰-50⁰ = 130⁰

⇒ x = 65⁰

∠B = ∠C = 65⁰

In Δ ABC since AB = BC ,so the triangle is isosceles

In an isosceles triangle the base angles are always equal

So ∠A = ∠C = x(Let’s assume)

Since sum of interior angles of a triangle is 180⁰

∠A + ∠B + ∠C = 180⁰

⇒ 2x = 180⁰-64⁰ = 116⁰

⇒ x = 58⁰

∠A = ∠C = 58⁰

Δ ABC is isosceles with AB = AC

∠BAC = 40⁰(Given)

In an isosceles triangle since the base angles are equal so

∠ABC = ∠ACB = y (Let’s assume)

∠ABC + ∠ACB + ∠BAC = 180⁰(Sum of interior angles of a triangle)

⇒ 2y = 180⁰-40⁰ = 140⁰

⇒ y = 70⁰

⇒ ∠ACB = 70⁰

∠ACB + x = 180⁰(Angle on a straight line)

⇒ x = 180⁰ - 70⁰ = 110⁰

Δ ACD is isosceles where AC = CD

∠DAC = ∠CDA = 30⁰

∠BAD = ∠DAC + ∠CAB

⇒ ∠BAD = 65⁰ + 30⁰ = 95⁰

In Δ ABD,

∠ABD + ∠BDA + ∠DAB = 180⁰ (Sum of interior angles of a triangle)

x = 180⁰ –(95⁰ + 30⁰) = 55⁰

Δ ABC is isosceles with AB = AC

∠ABC = ∠ACB = 55⁰ (Base angles are equal)

∠ABC + ∠ACB + ∠BAC = 180⁰(Sum of interior angles of a triangle)

⇒ ∠BAC = 180⁰-(55⁰ + 55⁰) = 70⁰

∠BAC-x = 180⁰-(55⁰ + 75⁰)

⇒ -x = 180⁰-200⁰ = -20⁰

⇒ x = 20⁰

Δ ABD and Δ ADC is isosceles

∠ABD = ∠BAD = 50⁰(Base angle of an isosceles triangle)

∠DAC = ∠DCA = x (Base angle of an isosceles triangle)

In Δ ABC

∠ABC + ∠ACB + ∠BAC = 180⁰(Sum of interior angles of a triangle)

50⁰ + 50⁰ + 2x = 180⁰

⇒ 2x = 80⁰

⇒ x = 40⁰

Δ ABC is an equilateral triangle

CD is drawn such that CD = BC

∠ACB = 60⁰(Interior angle of an equilateral triangle)

∠ACD = 180⁰-60⁰ = 120⁰(Angle on a straight line)

In an equilateral triangle since all sides are equal so

AC = CD

Hence Δ ACD is isosceles

Since base angles of an isosceles triangle is equal so

∠CAD = ∠CDA = x(Let us assume)

∠ACD + ∠CAD + ∠CDA = 180⁰(Sum of interior angles of a triangle)

⇒ 2x = 180⁰-120⁰ = 60⁰

⇒ x = 30⁰

∠ADC = 30⁰

Let Δ ABC be the isosceles triangle

BE and CF are the two perpendiculars drawn which cuts at O

In Δ BFC and Δ BEC

∠FBC = ∠ECB (Base angles of isosceles triangle)

∠BFC = ∠BEC = 90⁰ (Perpendiculars)

BC = BC(Common side)

So Δ BFC and Δ BEC are congruent to each other by R.H.S. axiom of congruency

BE = FC (Corresponding Part of Congruent Triangle)

Hence its proved

Δ ABC is isosceles with AB = AC

AD is the altitude on BC

In Δ ABD and Δ ACD

∠ABD = ∠ACD (Base angles of an isosceles triangle)

∠ADB = ∠ADC (AD is a perpendicular)

AD = AD (Common side)

So Δ ABD and Δ ACD are congruent to each other by A.A.S. axiom of congruency

BD = DC(Corresponding Parts of Congruent Triangles)

Since BD = DC so D is the midpoint of BC

So altitude AD bisects BC

Δ ABC is an equilateral triangle

Let AD be the perpendicular from A on BC

In Δ ABD and Δ ACD

AB = AC (ΔABC is equilateral)

∠ADB = ∠ADC (AD is a perpendicular)

AD = AD (Common side)

So Δ ABD and Δ ACD are congruent to each other by S.A.S. axiom of congruency

∠ABD = ∠ACD (Corresponding Parts of Congruent Triangles)

∠BAD = ∠CAD (Corresponding Parts of Congruent Triangles)

Since the triangle is equilateral and ∠ABD = ∠ACD, so

∠ABD = ∠ACD = ∠BAC

Hence the triangle is equiangular

In Δ DPC and Δ APB we have

DP = PB(P is the midpoint of BD)

∠DPC = ∠APB( Vertically Opposite)

∠DCP = ∠PAB (Alternate interior angle)

So Δ DPC and Δ APB are congruent to each other by A.A,S. axiom of congruency

In Δ ABC BE and CD are the perpendiculars drawn on sides AC and AB respectively

In Δ BDC and Δ BEC

∠BDC = ∠BEC = 90⁰ (Perpendiculars)

∠DBC = ∠ECB(Base angle of the isosceles triangle)

BC = BC(Common side)

So Δ BDC and Δ BEC are congruent to each other by A.A.S. axiom of congruency

DB = EC (Corresponding Part of Congruent Triangle) …(i)

AB = AC (Given) …(ii)

Subtracting (i) from (ii) we get

AB-DB = AC-EC

⇒ AD = AE

Hence Proved

In Δ AOP and Δ BOQ

AP = BQ (Given)

∠AOP = ∠BOQ (Vertically Opposite)

∠PAO = ∠OBO (Perpendiculars)

So AOP and Δ BOQ are congruent to each other by A.A.S. axiom of congruency

Hence we can say

AO = OB(Corresponding parts of Congruent triangles)

PO = OQ(Corresponding parts of Congruent triangles)

BD and CE are bisectors of ∠B and ∠C

∠ABD = ∠DBC

∠ACE = ∠BCE

Since Δ ABC is isosceles so

∠ABC = ∠ACB …(i)

Since BD and CE are bisectors so

2∠DBC = 2∠ECB

⇒ ∠DBC = ∠ECB …(ii)

BC = BC (Common) …(iii)

From (i) , (ii) and (iii) we can say

Δ BCE and Δ BCD are congruent to each other by A.A.S. axiom of congruency

So we can say

BD = EC (Corresponding parts of Congruent triangles)

Δ ABC is an equiangular triangle

Let AD be the perpendicular from A on BC

In Δ ABD and Δ ACD

∠ABD = ∠ACD (ΔABC is equiangular)

∠ADB = ∠ADC (AD is a perpendicular)

AD = AD (Common side)

So Δ ABD and Δ ACD are congruent to each other by A.A.S. axiom of congruency

AB = AC (Corresponding Parts of Congruent Triangles)

BD = DC(Corresponding Parts of Congruent Triangles)

Since the triangle is equiangular and AB = AC , so

AB = AC = BC

Hence the triangle is equilateral

In ΔADC and ΔADB

AC = AB(Given)

BD = DC(AD bisects BC)

AD = AD(Common)

So ΔADC ≅ ΔADB by S.S.S. axiom of congruency

In ΔPOQ, ΔQOR , ΔROS, and ΔSOP

PQ = QR = RS = SP (All sides of a square are equal)

∠POQ = ∠QOR = ∠ROS = ∠SOP = 90⁰ (Diagonals of a square bisect at right angle)

PO = QO = RO = SO (Diagonals bisect each other)

So ΔPOQ ≅ ΔQOR ≅ ΔROS ≅ ΔSOP by S.A.S. axiom of congruency

(i) In Δ ADB and Δ PQS

AB = PQ

AD = PS

BC = QR

Since D and S are midpoints of BC and QR

⇒ 2DC = 2SR

⇒ DC = SR

So Δ ADB and Δ PQS are congruent to each other by S.S.S. axiom of congruency

(ii) In Δ ADC and Δ PSR

AD = PS

BC = QR

Since D and S are midpoints of BC and QR

⇒ 2BD = 2QS

⇒ BD = QS

∠ADB = ∠PSO(Corresponding parts of Congruent triangles)

⇒ 180⁰-∠ADB = 180⁰-∠PSO

∠ADC = ∠PSR

So Δ ADC and Δ PSR are congruent to each other by S.A.S. axiom of congruency

Yes it follows Δ ABC and Δ PQR are congruent because Δ ABC is the sum of Δ ADB and Δ ADC and Δ PQR is the sum of Δ PQS and Δ PSR

In Δ LNP and Δ MNR

LP = MR( Since PQ = QR)

PN = NR(N is a midpoint)

∠LPN = ∠MRN(Base angles of an isosceles triangle)

So Δ LNP and Δ MNR to each other by S.A.S. axiom of congruency

Hence LN = MN (Corresponding parts of Congruent Triangles)

ABCD is a rectangle

AC is a diagonal

In Δ ABC and Δ ADC

AD = BC (Opposite sides of a rectangle)

∠ADC = ∠ABC = 90⁰(Angle of a rectangle)

AC = AC (Common side)

So Δ ABC and Δ ADC is congruent by R.H.S. axiom of congruency.

In Δ ABC, D is the midpoint on BC

Let PD and QD be the two perpendiculars drawn from D on AB and AC respectively

In Δ BPD and Δ CQD we have

PD = QD(Given)

∠DPB = ∠DQC = 90⁰(Perpendiculars)

BD = DC(D is a midpoint)

So Δ BPD and Δ CQD are congruent by R.H.S. axiom of congruency

So according to Corresponding Parts of Congruent triangles we get

∠PBD = ∠QCD

Since the two angles of ABC are equal so Δ ABC isosceles.

In Δ ABC BE and CF are the perpendiculars drawn on sides AC and AB respectively

Let O be the intersection of the two perpendiculars

In Δ BFC and Δ BEC

BE = FC(Given)

∠BFC = ∠BEC = 90⁰ (Perpendiculars)

BC = BC(Common side)

So Δ BFC and Δ BEC are congruent to each other by R.H.S. axiom of congruency

∠FBC = ∠ECB (Corresponding Part of Congruent Triangle)

Since two angles are the same so Δ ABC is isosceles.

The side opposite to the smallest angle is the smallest side and the angle opposite to the largest angle is the largest side

Since the sum of interior angles of a triangle is 180⁰

∠A + ∠B + ∠C = 180⁰

⇒ ∠A = 180⁰-(28° + 56°)

⇒ ∠A = 96⁰

Since ∠A is the largest angle so BC is the largest side

Since ∠B is the smallest angle so AC is the smallest side

The side opposite to the smallest angle is the smallest side and the angle opposite to the largest angle is the largest side

The largest side among the three sides of Δ ABC is CA = 7.6 cm

So the largest angle of Δ ABC is ∠B

The smallest side among the three sides of Δ ABC is AB = 4 cm

So the smallest angle of Δ ABC is ∠C

So in ascending order, the angles will be

∠C< ∠A < ∠B

In Δ ABC

AB = AD

∠B = 70⁰

∠ADB = 70⁰(Base angles of an isosceles triangle)

∠ADC = 180⁰-70⁰ = 110⁰(Angles on a straight line)

∠C = 40⁰

∠DAC = 180⁰-(110⁰ + 40⁰) = 30⁰

Comparing ∠ACD and ∠DAC we find

∠ACD>∠DAC

Since the side opposite to the largest angle is the largest so

AB > CD

ABCD is a quadrilateral with AD being the largest and BC being the smallest side

AC is a diagonal of the quadrilateral

Applying the law of Inequality of triangle in Δ BAC

∠BAC < ∠BCA (BC is the shortest side) …(i)

Applying the law of Inequality of triangle in Δ ACD

∠CAD < ∠DCA (AD is the largest side) …(ii)

Adding (i) and (ii) we get

∠BAC + ∠CAD < ∠BCA + ∠DCA

⇒ ∠A < ∠C

Hence Proved

According to triangle inequality the sum of smaller two sides is always greater than the largest side

In ΔAPB

AP + PB>AB …(i)

In ΔAPC

AP + PC>AC …(ii)

In ΔCPB

CP + PB>CB …(iii)

Adding (i),(ii) and (iii) we get

2 (PA + PB + PC)> AB + BC + CA

or changing the sign of inequality we can say

AB + BC + CA < 2 (PA + PB + PC)

(a) In the right triangle, the hypotenuse is the longest side.

Explanation: The longest side is the side opposite to the largest angle and in a right-angled triangle the largest angle is 90° and the side opposite to it is the hypotenuse.

(b) The sum of three altitudes of a triangle is less than its perimeter

Explanation: Consider ΔABC with altitudes as AP, BQ and CR on segments BC, AB and AC respectively

Consider ΔAPC

∠APC = 90° … AP is altitude

∠ACP is some acute angle less than 90°

Hence AC > AP … (i)

Similarly we can prove

BC > CR … (ii)

AB > BQ … (iii)

Adding (i), (ii) and (iii) we get

⇒ AC + BC + AB > AP + CR + BQ

⇒ sum of sides > sum of altitudes

(c) The sum of any two sides of a triangle is larger than the third side.

Explanation: if the sum of two sides is equal to the third then the points will be collinear and these points won’t form a triangle

(d) If two angles of a triangle are unequal, then the smaller angle has the smaller side opposite to it.

Explanation: The side opposite to larger angle is larger. And the side opposite to shorter angle is shorter. So, the smaller angle has smaller side opposite to it.

(e) The difference of any two sides of a triangle is less than the third side.

Explanation: if a, b, c are sides of a triangle we know that the sum of two sides is greater than the third

i.e. a + b > c

rearranging we get a > c – b

(f) If two sides of a triangle are unequal, then the larger side has larger angle opposite to it.

Explanation: consider ΔABC with AC the longer side than AB

Construct a line BD to AC from point B such that AB = AD

The angles are as marked

We have to prove that (x + z) > y

From figure

⇒ x + ∠BDC = 180° … linear pair of angles ∠ADB and ∠BDC

⇒ ∠BDC = 180° - x

⇒ z + y + ∠BDC = 180° … angles of ΔBDC

⇒ z + y +180° - x = 180°

⇒ z + y = x

⇒ x > y

⇒ ∠ABD >∠BDC

If x is greater than y then x plus something will obviously be greater than y

⇒ x + z > y

⇒ ∠ABC >∠ACB

Hence proved

Consider ΔABC with altitudes as AP, BQ and CR on segments BC, AB and AC respectively

Consider ΔAPC

∠APC = 90° … AP is altitude

∠ACP is some acute angle less than 90°

Hence AC > AP … (i)

Similarly we can prove

BC > CR … (ii)

AB > BQ … (iii)

Adding (i), (ii) and (iii) we get

⇒ AC + BC + AB > AP + CR + BQ

⇒ sum of sides > sum of altitudes

Consider ΔABC and AP is the median as shown BP = PC

Construct PQ such that AP = PQ

Consider ΔAPB and ΔQPC

AP = PQ … construction

∠APB = ∠QCP … vertically opposite angles

BP = PC … AP is median

Thus by SAS test for congruency

ΔAPB ≅ ΔQPC

⇒ AB = CQ … corresponding sides of congruent triangles…(i)

Consider ΔAQC

⇒ AC + CQ > AQ

Using (i)

⇒ AC + AB > AQ

But AQ = AP + PQ and AP = PQ by construction hence AQ = 2AP

⇒ AC + AB > 2AP

If a, b, c are sides of a triangle we know that the sum of two sides is greater than the third

i.e. a + b > c

rearranging we get a > c – b

In ΔABC and ΔDBC

BC = BC … common side

∠ABC = ∠DBC … given

AB = CD … given

Therefore, by SAS test for congruency

ΔABC ≅ ΔDCB

⇒ AC = BD … corresponding sides of congruent triangles

Hence proved AC = BD

In ΔAOB and ΔCOD

AO = CO … given

∠AOB = ∠COD … vertically opposite angles

BO = OD … given

Therefore, by SAS test for congruency

ΔAOB ≅ ΔCOD

⇒ AB = CD … corresponding sides of congruent triangles

Hence proved AB = CD

In ΔABC and ΔDCB

AB = BD … given

AC = CB … given

BC = BC … common side

Therefore, by SSS test for congruency

ΔABC ≅ ΔDBC

Hence proved

Consider ΔADQ and ΔDAP

DP = AQ … given

∠PAD = ∠QDA … angles of square both 90°

AD = AD … common side

Therefore, by SAS test for congruency

ΔADQ ≅ ΔDAP

⇒ AP = DQ … corresponding sides of congruent triangles (i)

Now AB and CD are sides of square therefore,

AB = CD

⇒ AP + PB = DQ + QC … from figure AB = AP + PB and CD = DQ + QC

But using (i) AP = DQ

⇒ DQ + PB = DQ + QC

⇒ PB = QC

⇒ BP = CQ

Hence proved

In ΔAPC and ΔAQB

AP = AQ … given

∠BAQ = ∠CAP … ∠A common angle

AB = AC … given

Therefore, by SAS test for congruency

ΔAPC ≅ ΔAQB

Hence proved

Let the isosceles triangle be ΔABC as shown with base as BC and equal sides are AB and AC

In an isosceles triangle the base angles are equal let them be ‘x°’

Thus ∠B = x° and ∠C = x°

Given that ∠A = 2 × (∠B + ∠C)

⇒ ∠A = 2 × (x° + x°)

⇒ ∠A = 2 × 2x°

⇒ ∠A = 4x°

Now the sum of all angles of a triangle is 180°

⇒ ∠A + ∠B + ∠C = 180°

⇒ 4x° + x° + x° = 180°

⇒ 6x° = 180°

⇒ x = 30°

Therefore, angles of triangle are ∠B = 30°, ∠C = 30° and ∠A = 4x = 4 × 30° = 120°

Consider ΔABC and AM is the bisector of ∠A and AM bisects the base BC so BM = CM

Extend segment AM to D such that AM = MD and join points C and D to from ΔDMC as shown

To prove ΔABC is isosceles we have to prove that AB = AC

Consider ΔAMB and ΔDMC

AM = MD … construction

∠AMB = ∠DMC … vertically opposite angles

BM = MC … AM bisects BC given

Hence by SAS test for congruency

ΔAMB ≅ ΔDMC

⇒ AB = CD … corresponding sides of congruent triangles … (i)

⇒ ∠BAM = ∠CDM … corresponding angles of congruent triangles … (a)

⇒ ∠BAM = ∠MAC … given AM is angle bisector of ∠A … (b)

Thus using (a) and (b) we can conclude that

∠CDM = ∠MAC … (c)

Now consider ΔACD

∠CAD = ∠CDA … from (c)

As the two angles are equal ΔACD is isosceles hence we can say that

⇒ AC = CD … (ii)

Now using (i) and (ii) we can conclude that

⇒ AB = AC

And hence ΔABC is isosceles triangle

The figure is as shown

As AB = AC and AB = AD thus AC = AD

Therefore, ΔACD is also isosceles triangle

Let the base angles of ΔABC be x and the base angles of ΔACD be y as shown

In order to prove ∠BCD is a right angle we have to prove that x + y = 90°

Consider ΔBCD

∠DBC = x,

∠DCB = x + y and

∠BDC = y

Sum of angles of a triangle is 180°

⇒ ∠DBC + ∠DCB + ∠BDC = 180°

⇒ x + x + y + y = 180°

⇒ 2x + 2y = 180°

⇒ 2(x + y) = 180°

⇒ x + y = 90°

⇒ ∠DCB = 90°

Hence proved ∠BCD is a right angle.

Given AB || CD and AD || BC hence ABCD is a parallelogram

⇒ AB = CD … opposite sides of parallelogram (i)

Consider ΔCED and ΔBEF

∠DEC = ∠BEF … vertically opposite angles

BE = EC … given

∠ECD = ∠EBF … alternate angles as AF || CD with transversal BC

Therefore, by ASA test for congruency

ΔCED ≅ ΔBEF

⇒ CD = BF … corresponding sides of congruent triangles

Using (i)

⇒ AB = BF

Hence proved