Triangles [Congruency in Triangles] — Exercise 9(A)

If △ ABC ≅ △ PQR, then

1. AC = PR

2. AC = PQ

3. BC = PR

4. BC = PQ

Answer

Since,

△ ABC ≅ △ PQR

We know that,

Corresponding parts of congruent triangles are equal.

∴ AC = PR (By C.P.C.T.C.)

Hence, Option 1 is the correct option.

Which of the following will hold true for the given figure :

1. AD = DC

2. CD = CB

3. ∠ACD ≠ ∠ACB

4. ∠B ≠ ∠D

Answer

From figure,

In △ ADC and △ ABC,

⇒ ∠DAC = ∠BAC (Given)

⇒ AD = AB (Given)

⇒ AC = AC (Common side)

∴ △ ADC ≅ △ ABC (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ CD = CB (By C.P.C.T.C.)

Hence, Option 2 is the correct option.

In the given figure, AM is the perpendicular bisector of BC. Then :

1. AB = AM

2. AC = BM

3. AB ≠ AC

4. AM bisects ∠BAC

Answer

From figure,

In △ ABM and △ ACM,

⇒ ∠AMB = ∠AMC (Both equal to 90°)

⇒ BM = MC (Since, AM is perpendicular bisector of BC)

⇒ AM = AM (Common side)

∴ △ ABM ≅ △ ACM (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ ∠BAM = ∠CAM (By C.P.C.T.C.)

∴ AM bisects ∠BAC.

Hence, Option 4 is the correct option.

Which of the following is true for the given figure:

1. ΔAPC ≅ ΔBPD

2. CP = DP

3. AB and CD bisect each other

4. all of the above are true.

Answer

In ΔAPC and ΔBPD,

⇒ ∠ACP = ∠BDP (Both are 90°)

⇒ ∠APC = ∠BPD (Vertically opposite angles are equal)

⇒ AC = BD (Given)

∴ ΔAPC ≅ ΔBPD (By AAS congruency criterion)

By C.P.C.T.,

⇒ CP = DP and AP = BP

Thus, AB and CD bisect each other.

∴ All the options are correct.

Hence, option 4 is the correct option.

In the following figure, ∠BAD = ∠EAC, BD = EC and ∠B = ∠E, then :

1. △ ABD ≇ △ AEC

2. △ ABC ≅ △ AED

3. △ ABC ≇ △ AED

4. △ ABD ≅ △ ADE

Answer

Given,

BD = EC = x (let)

∠B = ∠E

We know that,

Sides opposite to equal angles are equal.

∴ AE = AB

From figure,

BC = BD + DC = x + DC .......(1)

DE = DC + CE = DC + x .......(2)

From equations (1) and (2), we get :

BC = DE

In △ ABC and △ AED,

⇒ BC = DE (Proved above)

⇒ AB = AE (Proved above)

⇒ ∠B = ∠E (Given)

∴ △ ABC ≅ △ AED (By S.A.S. axiom)

Hence, Option 2 is the correct option.

Which of the following is true for the given figure :

1. △ ABD ≅ △ ACD

2. angle BAD ≠ angle CAD

3. △ ABD ≇ △ ACD

4. ∠EAB = ∠BAD

Answer

From figure,

⇒ ∠EAD = ∠FAD = 90°

⇒ ∠EAB = ∠FAC = x (let)

⇒ ∠BAD = ∠EAD - ∠EAB = 90° - x

⇒ ∠CAD = ∠FAD - ∠FAC = 90° - x.

∴ ∠BAD = ∠CAD.

In △ ABD and △ ACD,

⇒ AD = AD (Common side)

⇒ ∠ADB = ∠ADC (Both equal to 90°)

⇒ ∠BAD = ∠CAD (Proved above)

∴ △ ABD ≅ △ ACD (By A.S.A. axiom)

Hence, Option 1 is the correct option.

In the given figure, ∠x = ∠y and PO = RO, then :

1. RB = AO

2. BO = PA

3. BP = AR

4. RB = OB

Answer

From figure,

OP is a straight line.

∴ ∠OAR + x = 180°

⇒ ∠OAR = 180° - x

OR is a straight line.

∴ ∠PBO + y = 180°

⇒ ∠PBO = 180° - y

Since, ∠x = ∠y

∴ ∠OAR = ∠PBO

In △ PBO and △ OAR,

⇒ PO = RO (Given)

⇒ ∠PBO = ∠OAR (Proved above)

⇒ ∠O = ∠O (Common angle)

∴ △ PBO ≅ △ OAR (By A.A.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ BP = AR.

Hence, Option 3 is the correct option.

In the given figure, BC // DA and BC = DA, then :

1. AB and CD bisect each other

2. AB ≠ CD

3. OA = OC

4. OA = OD

Answer

In △ BOC and △ DOA,

⇒ ∠DOA = ∠BOC (Vertically opposite angles are equal)

⇒ ∠OAD = ∠OBC (Alternate angles are equal)

⇒ ∠ODA = ∠OCB (Alternate angles are equal)

∴ △ BOC ≅ △ DOA (By A.A.A. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ OD = OC and OA = OB.

∴ AB and CD bisect each other.

Hence, Option 1 is the correct option.

The following figure shows a circle with center O. If OP is perpendicular to AB, prove that AP = BP.

Answer

Join OA and OB.

In △ OAP and △ OBP,

⇒ OP = OP (Common side)

⇒ OA = OB (Radius of same circle)

⇒ ∠OPA = ∠OPB (Both equal to 90°)

∴ △ OAP ≅ △ OBP (By R.H.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ AP = BP.

Hence, proved that AP = BP.

In a triangle ABC, D is mid-point of BC; AD is produced upto E, so that DE = AD. Prove that :

(i) △ ABD and △ ECD are congruent.

(ii) AB = EC

(iii) AB is parallel to EC.

Answer

△ ABC with AD produced upto E is shown in the figure below:

(i) In △ ABD and △ ECD,

⇒ AD = DE (Given)

⇒ BD = DC (As D is the mid-point of BC)

⇒ ∠ADB = ∠CDE (Vertically opposite angles are equal)

∴ △ ABD ≅ △ ECD (By S.A.S. axiom)

Hence, proved that △ ABD ≅ △ ECD.

(ii) Since,

△ ABD ≅ △ ECD (Proved above)

We know that,

Corresponding parts of congruent triangles are equal.

∴ AB = EC.

Hence, proved that AB = EC.

(iii) From figure,

∠ABD = ∠DCE (By C.P.C.T.C.)

Since, these are alternate angles and are also equal,

thus, it can be said, AB and EC are parallel with AE as transversal.

Hence, proved that AB is parallel to EC.

From the given diagram, in which ABCD is a parallelogram, ABL is a line segment and E is mid point of BC.

Prove that :

(i) △ DCE ≅ △ LBE

(ii) AB = BL

(iii) AL = 2DC

Answer

(i) In parallelogram ABCD,

AB || CD

Since, ABL is a straight line.

∴ BL || CD

In △ LBE and △ DCE,

⇒ BE = CE (Given)

⇒ ∠BEL = ∠CED (Vertically opposite angles are equal)

⇒ ∠CDE = ∠ELB (Alternate angles are equal)

∴ △ LBE ≅ △ DCE (By A.A.S. axiom)

Hence, proved that △ LBE ≅ △ DCE.

(ii) We know that,

Opposite sides of parallelogram are equal.

∴ AB = CD .......(1)

Since, △ LBE ≅ △ DCE

We know that,

Corresponding parts of congruent triangles are equal.

∴ DC = BL .......(2)

From equation (1) and (2), we get :

AB = BL.

Hence, proved that AB = BL.

(iii) Given,

⇒ AB = CD

⇒ AL - BL = CD

⇒ AL - CD = CD (Using Eq 2)

⇒ AL = CD + CD

⇒ AL = 2CD.

Hence, proved that AL = 2CD.

On the sides AB and AC of triangle ABC, equilateral triangles ABD and ACE are drawn. Prove that :

(i) ∠CAD = ∠BAE

(ii) CD = BE.

Answer

△ ABC with equilateral triangles ABD and ACE drawn on its sides AB and AC, respectively are is shown below:

(i) Since, ABD and ACE are equilateral triangles.

∴ ∠BAD = ∠CAE (Both equal to 60°)

Adding ∠CAB on both sides we get :

⇒ ∠BAD + ∠CAB = ∠CAE + ∠CAB

⇒ ∠CAD = ∠BAE.

Hence, proved that ∠CAD = ∠BAE.

(ii) In △ CAD and △ BAE,

⇒ AC = AE (△ ACE is equilateral triangle)

⇒ ∠CAD = ∠BAE (Proved above)

⇒ AD = AB (△ ABD is equilateral triangle)

∴ △ CAD ≅ △ BAE (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ CD = BE.

Hence, proved that CD = BE.

A line segment AB is bisected at point P and through point P another line segment PQ, which is perpendicular to AB, is drawn. Show that : QA = QB.

Answer

In △ QAP and △ QBP,

⇒ QP = QP (Common side)

⇒ ∠QPA = ∠QPB (Both equal to 90°)

⇒ AP = PB (Since, AB is bisected at point P)

∴ △ QAP ≅ △ QBP (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ QA = QB.

Hence, proved that QA = QB.

In the following diagrams, ABCD is a square and APB is an equilateral triangle. In each case,

(i) Prove that : △ APD ≅ △ BPC

(ii) Find the angles of △ DPC.

Answer

For 1st case :

(i) We know that,

Each interior angle in a square is 90° and in an equilateral triangle is 60°.

From figure,

⇒ ∠DAP = ∠DAB - ∠PAB = 90° - 60° = 30°.

⇒ ∠CBP = ∠CBA - ∠PBA = 90° - 60° = 30°.

In △ APD and △ BPC,

⇒ ∠DAP = ∠CBP (Proved above)

⇒ AP = PB (Since, APB is an equilateral triangle)

⇒ AD = BC (Since, ABCD is a square)

∴ △ APD ≅ △ BPC (By S.A.S. axiom)

Hence, proved that △ APD ≅ △ BPC.

(ii) We know that,

⇒ AP = AB (Since, APB is an equilateral triangle)

⇒ AB = AD (Since, ABCD is a square)

∴ AP = AD

Thus, APD is an isosceles triangle.

We know that,

Angles opposite to equal sides are equal.

∴ ∠APD = ∠ADP = x (let)

In △ APD,

By angle sum property of triangle,

⇒ ∠APD + ∠ADP + ∠DAP = 180°

⇒ x + x + 30° = 180°

⇒ 2x + 30° = 180°

⇒ 2x = 180° - 30°

⇒ 2x = 150°

⇒ x = $\dfrac{150°}{2}$ = 75°

⇒ ∠APD = ∠ADP = 75°.

From figure,

⇒ ∠ADC = 90° (Interior angle of a square is 90°)

⇒ ∠CDP = ∠ADC - ∠ADP = 90° - 75° = 15°.

Since, △ APD ≅ △ BPC,

⇒ DP = PC (By C.P.C.T.C.)

∴ DPC is an isosceles triangle.

⇒ ∠CDP = ∠DCP = 15° (Angles opposite to equal sides are equal)

In △ DCP,

By angle sum property of triangle,

⇒ ∠DCP + ∠CDP + ∠DPC = 180°

⇒ 15° + 15° + ∠DPC = 180°

⇒ 30° + ∠DPC = 180°

⇒ ∠DPC = 180° - 30° = 150°.

Hence, ∠CDP = ∠DCP = 15° and ∠DPC = 150°.

For 2nd case :

(i) We know that,

Each interior angle in a square is 90° and each interior angle in an equilateral triangle is 60°.

From figure,

⇒ ∠DAP = ∠DAB + ∠BAP

⇒ ∠DAP = 90° + 60° = 150°.

⇒ ∠CBP = ∠CBA + ∠ABP

⇒ ∠CBP = 90° + 60° = 150°.

In △ APD and △ BPC,

⇒ ∠DAP = ∠CBP (Both equal to 150°)

⇒ AP = PB (Since, APB is an equilateral triangle)

⇒ AD = BC (Since, ABCD is a square)

∴ △ APD ≅ △ BPC (By S.A.S. axiom)

(ii) ABCD is a square.

∴ AB = AD = DC = BC

APB is an equilateral triangle.

∴ AP = PB = AB

So, we get :

AP = AD and PB = BC

∴ △ APD and △ BPC are isosceles triangle.

We know that,

Angles opposite to equal sides are equal sides are equal.

∴ ∠APD = ∠ADP = x (let) and ∠BPC = ∠BCP = y (let)

In △ APD,

By angle sum property of triangle,

⇒ ∠APD + ∠ADP + ∠DAP = 180°

⇒ x + x + 150° = 180°

⇒ 2x + 150° = 180°

⇒ 2x = 180° - 150°

⇒ 2x = 30°

⇒ x = $\dfrac{30°}{2}$ = 15°.

In △ BPC,

By angle sum property of triangle,

⇒ ∠BPC + ∠BCP + ∠PBC = 180°

⇒ y + y + 150° = 180°

⇒ 2y + 150° = 180°

⇒ 2y = 180° - 150°

⇒ 2y = 30°

⇒ y = $\dfrac{30°}{2}$ = 15°.

From figure,

⇒ ∠PDC = ∠ADC - ∠ADP = 90° - 15° = 75°.

⇒ ∠PCD = ∠BCD - ∠BCP = 90° - 15° = 75°.

In △ DPC,

By angle sum property of triangle,

⇒ ∠DPC + ∠PDC + ∠PCD = 180°

⇒ ∠DPC + 75° + 75° = 180°

⇒ ∠DPC + 150° = 180°

⇒ ∠DPC = 180° - 150° = 30°.

Hence, ∠DPC = 30° and ∠PDC = ∠PCD = 75°.

In the figure, given below, triangle ABC is right-angled at B. ABPQ and ACRS are squares. Prove that :

(i) △ ACQ and △ ASB are congruent.

(ii) CQ = BS.

Answer

From figure,

⇒ ∠QAC = ∠QAB + ∠BAC = 90° + ∠BAC.

⇒ ∠BAS = ∠CAS + ∠BAC = 90° + ∠BAC.

∴ ∠QAC = ∠BAS.

In △ QAC and △ BAS,

⇒ QA = AB (Since, ABPQ is a square)

⇒ ∠QAC = ∠BAS (Proved above)

⇒ AC = AS (Since, ACRS is a square)

∴ △ QAC ≅ △ BAS (By S.A.S. axiom)

Hence, proved that △ ACQ and △ ASB are congruent.

(ii) Since, △ QAC ≅ △ BAS

We know that,

Corresponding parts of congruent triangles are equal.

∴ CQ = BS.

Hence, proved that CQ = BS.

In a △ ABC, BD is the median to the side AC, BD is produced to E such that BD = DE. Prove that : AE is parallel to BC.

Answer

△ ABC with BD as median to the side AC and BD is produced to E such that BD = DE is shown below:

In △ ADE and △ BDC,

⇒ ∠ADE = ∠BDC (Vertically opposite angles are equal)

⇒ AD = DC (As BD is median to side AC)

⇒ BD = DE (Given)

∴ △ ADE ≅ △ BDC (By S.A.S. axiom).

We know that,

Corresponding parts of congruent triangles are equal.

∴ ∠EAD = ∠DCB

The above angles are alternate angles, since they are equal,

∴ AE || BC.

Hence, proved that AE is parallel to BC.

ABCD is a parallelogram. The sides AB and AD are produced to E and F respectively such that AB = BE and AD = DF. Prove that :

△ BEC ≅ △ DCF

Answer

Parallelogram ABCD with sides AB and AD are produced to E and F, respectively are is shown below:

Given,

AD = DF ..........(1)

AB = BE ..........(2)

We know that,

Opposite sides of parallelogram are equal.

∴ AD = BC ........(3)

and,

AB = CD ..........(4)

From equations (1) and (3), we get :

⇒ BC = DF

From equations (2) and (4), we get :

⇒ BE = CD

We know that,

Opposite angles of a parallelogram are equal.

∴ ∠ABC = ∠ADC = x (let)

From figure,

Since, AE is a straight line.

⇒ ∠CBE + ∠ABC = 180°

⇒ ∠CBE + x = 180°

⇒ ∠CBE = 180° - x.

Since, AF is a straight line.

⇒ ∠CDF + ∠ADC = 180°

⇒ ∠CDF + x = 180°

⇒ ∠CDF = 180° - x.

In △ BEC and △ DCF,

⇒ BE = CD (Proved above)

⇒ BC = DF (Proved above)

⇒ ∠CBE = ∠CDF (Both equal to 180° - x)

∴ △ BEC ≅ △ DCF (By S.A.S. axiom).

Hence proved that △ BEC ≅ △ DCF.

In the following figure, ABC is an equilateral triangle in which QP is parallel to AC. Side AC is produced upto point R so that CR = BP.

Prove that QR bisects PC.

Answer

Given,

∆ ABC is an equilateral triangle.

∴ ∠ABC = ∠BCA = ∠CAB = 60° (Each interior angle of equilateral triangle equals to 60°.)

From figure,

⇒ ∠BPQ = ∠BCA = 60° (Corresponding angles are equal)

⇒ ∠BQP = ∠BAC = 60° (Corresponding angles)

In ∆BPQ,

By angle sum property of triangle,

⇒ ∠BQP + ∠BPQ + ∠QBP = 180°

⇒ 60° + 60° + ∠QBP = 180°

⇒ 120° + ∠QBP = 180°

⇒ ∠QBP = 180° - 120° = 60°.

Since, all the interior angles = 60°.

∴ △ BPQ is an equilateral triangle i.e., BP = PQ = BQ.

Given,

BP = CR

Since, BP = PQ.

∴ PQ = CR.

In △ MPQ and △ MCR,

⇒ ∠PQM = ∠MRC (Alternate interior angles are equal)

⇒ ∠PMQ = ∠CMR (Vertically opposite angles are equal)

⇒ PQ = CR (Proved above)

∴ ∆ MPQ ≅ ∆ MCR (By A.A.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ MP = MC

Thus, QR bisects PC.

Hence, proved that QR bisects PC.

PQRS is a parallelogram. L and M are points on PQ and SR respectively such that PL = MR. Show that LM and QS bisect each other.

Answer

Parallelogram PQRS is shown in the figure below:

We know that,

Opposite angles of a parallelogram are equal.

∠Q = ∠S = x (let).

Diagonals of a parallelogram bisect the interior angles.

∴ QS bisects interior angles Q and S.

∴ ∠LQN = $\dfrac{∠Q}{2} = \dfrac{x}{2}$ and ∠NSM = $\dfrac{∠S}{2} = \dfrac{x}{2}$.

∴ ∠LQN = ∠NSM.

We know that,

Opposite sides of a parallelogram are equal.

∴ PQ = SR = a (let)

Given,

⇒ PL = MR = b (let)

From figure,

⇒ LQ = PQ - PL = a - b

⇒ MS = SR - MR = a - b

∴ LQ = MS.

In △ LNQ and △ MNS,

⇒ ∠LNQ = ∠MNS (Vertically opposite angles are equal)

⇒ LQ = MS (Proved above)

⇒ ∠LQN = ∠NSM (Proved above)

∴ △ LNQ ≅ △ MNS (By A.A.S. axiom).

We know that,

Corresponding parts of congruent triangles are equal.

∴ QN = NS and LN = NM.

Hence, proved that LM and QS bisect each other at point of intersection.