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

In the given figure, AB = PQ, BC = QR and median AM = median PN, then :

1. AC ≠ PR

2. BM ≠ QN

3. △ ABM ≇ △ PQN

4. △ ABC ≅ △ PQR

Answer

Given,

AM and PN are medians of triangle ABC and PQR.

∴ BM = MC and QN = NR.

BC = QR (Given)

⇒ $\dfrac{BC}{2} = \dfrac{QR}{2}$

⇒ BM = QN.

In △ ABM and △ PQN,

⇒ AB = PQ (Given)

⇒ AM = PN (Given)

⇒ BM = QN (Proved above)

∴ △ ABM ≅ △ PQN (By S.S.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ ∠AMB = ∠PNQ (By C.P.C.T.C.)

⇒ 180° - ∠AMB = 180° - ∠PNQ

⇒ ∠AMC = ∠PNR

Given,

⇒ BC = QR

⇒ $\dfrac{BC}{2} = \dfrac{QR}{2}$

⇒ MC = NR.

In △ AMC and △ PNR,

⇒ AM = PN (Given)

⇒ ∠AMC = ∠PNR (Proved above)

⇒ MC = NR (Proved above)

∴ △ AMC ≅ △ PNR (By S.A.S. axiom)

∴ AC = PR (By C.P.C.T.C.)

In △ ABC and △ PQR,

⇒ AB = PQ (Given)

⇒ BC = QR (Given)

⇒ AC = PR (Proved above)

∴ △ ABC ≅ △ PQR (By S.S.S. axiom)

Hence, Option 4 is the correct option.

In triangles ABC and DEF, AB = DE and AC = EF, then to make these two triangles congruent, we must have :

1. BC = DF

2. ∠A = ∠E

3. any of (1) and (2)

4. none of (1) and (2)

Answer

Given,

In triangle ABC and DEF,

AB = DE (Given)

AC = EF (Given)

According to option 1 :

BC = DF

In such case all three sides are equal.

Then both the triangles will be congruent by S.S.S. axiom.

According to option 2 :

∠A = ∠E

In such case two sides and the included angle will be equal.

Then both the triangles will be congruent by S.A.S. axiom.

Hence, option 3 is the correct option.

In quadrilateral ABCD, AB = AC and BD = CD, then AD bisects :

1. angle ADC

2. angle BAD

3. angle BAC

4. angle ABC

Answer

In △ BAD and △ CAD,

⇒ AB = AC (Given)

⇒ AD = AD (Common side)

⇒ BD = CD (Given)

∴ △ BAD ≅ △ CAD (By S.S.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ ∠BAD = ∠CAD.

∴ AD bisects ∠BAC.

Hence, Option 3 is the correct option.

The given figure shows a circle with center O. P is mid-point of chord AB. Show that OP is perpendicular to AB.

Answer

Join OA and OB.

In △ OAP and △ OBP,

⇒ OP = OP (Common side)

⇒ OA = OB (Radius of same circle)

⇒ AP = PB (Since, P is mid-point of chord AB)

∴ △ OAP ≅ △ OBP (By S.S.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ ∠OPA = ∠OPB = x (let)

Since, AB is a straight line.

∴ ∠OPA + ∠OPB = 180°

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = $\dfrac{180°}{2}$ = 90°.

Hence, proved that OP is perpendicular to chord AB.

A triangle ABC has ∠B = ∠C. Prove that :

(i) the perpendiculars from the mid-point of BC to AB and AC are equal.

(ii) the perpendicular from B and C to the opposite sides are equal.

Answer

△ ABC is shown in the figure below:

(i) From figure,

In △ BDE and △ CDF,

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

⇒ ∠B = ∠C (Given)

⇒ ∠DEB = ∠CFD (Both equal to 90°)

∴ △ BDE ≅ △ CDF (By A.A.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ DE = DF.

Hence, proved that the perpendiculars from the mid-point of BC to AB and AC are equal.

(ii) Let perpendiculars from B and C touch sides AC and AB at point H and G respectively.

In triangle ABC,

⇒ ∠B = ∠C

⇒ AC = AB (Sides opposite to equal angles in a triangle are equal)

From figure,

In △ ABH and △ ACG,

⇒ ∠AHB = ∠AGC (Both equal to 90°)

⇒ ∠BAH = ∠CAG (Common angle)

⇒ AB = AC (Proved above)

∴ △ ABH ≅ △ ACG (By A.A.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ BH = GC.

Hence, proved that the perpendicular from B and C to the opposite sides are equal.

In the adjoining figure, QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR. If XS ⊥ QR and XT ⊥ PQ; prove that :

(i) △ XTQ ≅ △ XSQ

(ii) PX bisects angle P.

Answer

(i) In △ XTQ and △ XSQ,

⇒ ∠XSQ = ∠XTQ (Both equal to 90°)

⇒ XQ = XQ (Common side)

⇒ ∠XQT = ∠XQS (Since, XQ is the bisector of ∠Q)

∴ △ XTQ ≅ △ XSQ (By A.A.S. axiom)

Hence, proved that △ XTQ ≅ △ XSQ.

(ii) Draw a perpendicular from X on PR i.e. XU.

In △ XSR and △ XUR,

⇒ ∠XSR = ∠XUR (Both are equal to 90°)

⇒ ∠XRS = ∠XRU (As XR is bisector of ∠R)

⇒ XR = XR (Common side)

∴ △ XSR ≅ △ XUR (By A.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ XU = XS ......(1)

As, △ XTQ ≅ △ XSQ

∴ XS = XT ......(2)

In △ XUP and △ XTP,

From (1) and (2) we get,

⇒ XU = XT

⇒ XP = XP (Common)

⇒ ∠XTP = ∠XUP (Both are equal to 90°)

∴ △XUP ≅ △XTP by RHS axiom.

We know that,

Corresponding parts of congruent triangle are equal.

∴ ∠XPU = ∠XPT

Hence, proved that PX is bisector of ∠P.

In the following figures, the sides AB and BC and the median AD of the triangle ABC are respectively equal to the sides PQ and QR and median PS of the triangle PQR. Prove that △ ABC and △ PQR are congruent.

Answer

Given,

BC = QR = x (let)

AD and PS are median of triangle ABC and PQR.

∴ BD = $\dfrac{1}{2}BC = \dfrac{x}{2}$

and

QS = $\dfrac{1}{2}QR = \dfrac{x}{2}$.

In △ ABD and △ PQS,

⇒ AB = PQ (Given)

⇒ AD = PS (Given)

⇒ BD = QS (Proved above)

∴ △ ABD ≅ △ PQS (By S.S.S. axiom).

We know that,

Corresponding parts of congruent triangle are equal.

∴ ∠B = ∠Q.

In △ ABC and △ PQR,

⇒ AB = PQ (Given)

⇒ BC = QR (Given)

⇒ ∠B = ∠Q (Proved above)

∴ △ ABC ≅ △ PQR (By S.A.S. axiom).

Hence, proved that △ ABC ≅ △ PQR.

In the following figure, OA = OC and AB = BC. Prove that :

(i) ∠AOB = 90°

(ii) △ AOD ≅ △ COD

(iii) AD = CD

Answer

(i) In △ AOB and △ COB,

⇒ OA = OC (Given)

⇒ AB = BC (Given)

⇒ OB = OB (Common side)

∴ △ AOB ≅ △ COB (By S.S.S. axiom).

We know that,

Corresponding parts of congruent triangles are equal.

∴ ∠AOB = ∠COB = x (let)

From figure,

AC is a straight line.

∴ ∠AOB + ∠COB = 180°

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = $\dfrac{180}{2}$ = 90°.

∴ ∠AOB = 90°.

Hence, proved that ∠AOB = 90°.

(ii) We know that,

Vertically opposite angles are equal.

∴ ∠AOD = ∠COB = 90° and ∠COD = ∠AOB = 90°.

In △ AOD and △ COD,

⇒ OA = OC (Given)

⇒ ∠AOD = ∠COD (Both equal to 90°)

⇒ OD = OD (Common side)

∴ △ AOD ≅ △ COD (By S.A.S. axiom).

Hence, proved that △ AOD ≅ △ COD.

(iii) Since, △ AOD ≅ △ COD.

We know that,

Corresponding parts of congruent triangles are equal.

∴ AD = CD

Hence, proved that AD = CD.

The following figure shows a triangle ABC in which AB = AC. M is a point on AB and N is a point on AC such that BM = CN. Prove that :

(i) AM = AN

(ii) △ AMC ≅ △ ANB

(iii) BN = CM

(iv) △ BMC ≅ △ CNB

Answer

(i) Given,

AB = AC = x (let) and BM = CN = y (let)

From figure,

⇒ AM = AB - BM = x - y

⇒ AN = AC - CN = x - y

∴ AM = AN.

Hence, proved that AM = AN.

(ii) In △ AMC and △ ANB,

⇒ AM = AN (Proved above)

⇒ ∠MAC = ∠NAB (Common angle)

⇒ AC = AB (Given)

∴ △ AMC ≅ △ ANB (By S.A.S. axiom).

Hence, proved that △ AMC ≅ △ ANB.

(iii) We know that,

Corresponding parts of congruent triangles are equal.

Since,

△ AMC ≅ △ ANB

∴ CM = BN.

Hence, proved that BN = CM.

(iv) We know that,

Angles opposite to equal sides are equal.

Since,

AB = AC

∴ ∠C = ∠B.

In △ BMC and △ CNB,

⇒ BM = CN (Given)

⇒ BC = BC (Common side)

⇒ ∠B = ∠C (Proved above)

∴ △ BMC ≅ △ CNB (By S.A.S. axiom).

Hence, proved that △ BMC ≅ △ CNB.

In a triangle ABC, AB = BC, AD is perpendicular to side BC and CE is perpendicular to side AB. Prove that : AD = CE.

Answer

△ ABC is shown below:

In △ ABD and △ CBE,

⇒ ∠B = ∠B (Common angle)

⇒ ∠ADB = ∠CEB (Both equal to 90°)

⇒ AB = BC (Given)

∴ △ ABD ≅ △ CBE (By A.A.S. axiom).

We know that,

Corresponding parts of congruent triangles are equal.

∴ AD = CE

Hence, proved that AD = CE.