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**Exercise 7.3****1. Î”ABC and Î”DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that (i) Î”ABD â‰… Î”ACD (ii) Î”ABP â‰… Î”ACP (iii) AP bisects âˆ A as well as âˆ D. (iv) AP is the perpendicular bisector of BC. **

**Answer**

Given,

Î”ABC and Î”DBC are two isosceles triangles.

(i) In Î”ABD and Î”ACD,

AD = AD (Common)

AB = AC (Î”ABC is isosceles)

BD = CD (Î”DBC is isosceles)

Therefore, Î”ABD â‰… Î”ACD by SSS congruence condition.

(ii) In Î”ABP and Î”ACP,

AP = AP (Common)

âˆ PAB = âˆ PAC (Î”ABD â‰… Î”ACD so by CPCT)

AB = AC (Î”ABC is isosceles)

Therefore, Î”ABP â‰… Î”ACP by SAS congruence condition.

(iii) âˆ PAB = âˆ PAC by CPCT as Î”ABD â‰… Î”ACD.

AP bisects âˆ A. --- (i)

also,

In Î”BPD and Î”CPD,

PD = PD (Common)

BD = CD (Î”DBC is isosceles.)

BP = CP (Î”ABP â‰… Î”ACP so by CPCT.)

Therefore, Î”BPD â‰… Î”CPD by SSS congruence condition.

Thus, âˆ BDP = âˆ CDP by CPCT. --- (ii)

By (i) and (ii) we can say that AP bisects âˆ A as well as âˆ D.

(iv) âˆ BPD = âˆ CPD (by CPCT as Î”BPD â‰… Î”CPD)

and BP = CP --- (i)

also,

âˆ BPD + âˆ CPD = 180Â° (BC is a straight line.)

â‡’ 2âˆ BPD = 180Â°

â‡’ âˆ BPD = 90Â° ---(ii)

From (i) and (ii),

AP is the perpendicular bisector of BC.**2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC **

**(ii) AD bisects âˆ A.****Answer**

Given,

AD is an altitude and AB = AC

(i) In Î”ABD and Î”ACD,

âˆ ADB = âˆ ADC = 90Â°

AB = AC (Given)

AD = AD (Common)

Therefore, Î”ABD â‰… Î”ACD by RHS congruence condition.

Now,

BD = CD (by CPCT)

Thus, AD bisects BC

(ii) âˆ BAD = âˆ CAD (by CPCT)

Thus, AD bisects âˆ A.**3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Î”PQR (see Fig. 7.40). Show that: (i) Î”ABM â‰… Î”PQN (ii) Î”ABC â‰… Î”PQR**

**Answer**

Given,

AB = PQ, BC = QR and AM = PN

(i) 1/2 BC = BM and 1/2QR = QN (AM and PN are medians)

also,

BC = QR

â‡’ 1/2 BC = 1/2QR

â‡’ BM = QN

In Î”ABM and Î”PQN,

AM = PN (Given)

AB = PQ (Given)

BM = QN (Proved above)

Therefore, Î”ABM â‰… Î”PQN by SSS congruence condition.

(ii) In Î”ABC and Î”PQR,

AB = PQ (Given)

âˆ ABC = âˆ PQR (by CPCT)

BC = QR (Given)

Therefore, Î”ABC â‰… Î”PQR by SAS congruence condition.**4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles. Answer **

Given,

BE and CF are two equal altitudes.

In Î”BEC and Î”CFB,

âˆ BEC = âˆ CFB = 90Â° (Altitudes)

BC = CB (Common)

BE = CF (Common)

Therefore, Î”BEC â‰… Î”CFB by RHS congruence condition.

Now,

âˆ C = âˆ B (by CPCT)

Thus, AB = AC as sides opposite to the equal angles are equal.**5. ABC is an isosceles triangle with AB = AC. Draw AP âŠ¥ BC to show that âˆ B = âˆ C. Answer**

Given,

AB = AC

In Î”ABP and Î”ACP,

âˆ APB = âˆ APC = 90Â° (AP is altitude)

AB = AC (Given)

AP = AP (Common)

Therefore, Î”ABP â‰… Î”ACP by RHS congruence condition.

Thus, âˆ B = âˆ C (by CPCT)**Exercise 7.4****1. Show that in a right angled triangle, the hypotenuse is the longest side. Answer**

ABC is a triangle right angled at B.

Now,

âˆ A + âˆ B + âˆ C = 180Â°

â‡’ âˆ A + âˆ C = 90Â° and âˆ B is 90Â°.

Since, B is the largest angle of the triangle, the side opposite to it must be the largest.

So, BC is the hypotenuse which is the largest side of the right angled triangle ABC.**2. In Fig. 7.48, sides AB and AC of Î”ABC are extended to points P and Q respectively. Also, âˆ PBC < âˆ QCB. Show that AC > AB.**

**Answer**

Given,

âˆ PBC < âˆ QCB

Now,

âˆ ABC + âˆ PBC = 180Â°

â‡’ âˆ ABC = 180Â° - âˆ PBC

also,

âˆ ACB + âˆ QCB = 180Â°

â‡’ âˆ ACB = 180Â° - âˆ QCB

Since,

âˆ PBC < âˆ QCB therefore, âˆ ABC > âˆ ACB

Thus, AC > AB as sides opposite to the larger angle is larger.

**3. In Fig. 7.49, âˆ B < âˆ A and âˆ C < âˆ D. Show that AD < BC.**

**Answer**

Given,

âˆ B < âˆ A and âˆ C < âˆ D

Now,

AO < BO --- (i) (Side opposite to the smaller angle is smaller)

OD < OC ---(ii) (Side opposite to the smaller angle is smaller)

Adding (i) and (ii)

AO + OD < BO + OC

â‡’ AD < BC**4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50). Show that âˆ A > âˆ C and âˆ B > âˆ D.**

**Answer**

In Î”ABD,

AB < AD < BD

âˆ´ âˆ ADB < âˆ ABD --- (i) (Angle opposite to longer side is larger.)

Now,

In Î”BCD,

BC < DC < BD

âˆ´ âˆ BDC < âˆ CBD --- (ii)

Adding (i) and (ii) we get,

âˆ ADB + âˆ BDC < âˆ ABD + âˆ CBD

â‡’ âˆ ADC < âˆ ABC

â‡’ âˆ B > âˆ D

Similarly,

In Î”ABC,

âˆ ACB < âˆ BAC --- (iii) (Angle opposite to longer side is larger.)

Now,

In Î”ADC,

âˆ DCA < âˆ DAC --- (iv)

Adding (iii) and (iv) we get,

âˆ ACB + âˆ DCA < âˆ BAC + âˆ DAC

â‡’ âˆ BCD < âˆ BAD

â‡’ âˆ A > âˆ C

**5. In Fig 7.51, PR > PQ and PS bisects âˆ QPR. Prove that âˆ PSR > âˆ PSQ.**

**Answer**

Given,

PR > PQ and PS bisects âˆ QPR

To prove,

âˆ PSR > âˆ PSQ

Proof,

âˆ PQR > âˆ PRQ --- (i) (PR > PQ as angle opposite to larger side is larger.)

âˆ QPS = âˆ RPS --- (ii) (PS bisects âˆ QPR)

âˆ PSR = âˆ PQR + âˆ QPS --- (iii) (exterior angle of a triangle equals to the sum of opposite interior angles)

âˆ PSQ = âˆ PRQ + âˆ RPS --- (iv) (exterior angle of a triangle equals to the sum of opposite interior angles)

Adding (i) and (ii)

âˆ PQR + âˆ QPS > âˆ PRQ + âˆ RPS

â‡’ âˆ PSR > âˆ PSQ [from (i), (ii), (iii) and (iv)]**6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.****Answer**

Let* l *is a line segment and B is a point lying o it. We drew a line AB perpendicular to* l*. Let C be any other point on *l*.

To prove,

AB < AC

Proof,

In Î”ABC,

âˆ B = 90Â°

Now,

âˆ A + âˆ B + âˆ C = 180Â°

â‡’ âˆ A + âˆ C = 90Â°

âˆ´ âˆ C mustbe acute angle. or âˆ C < âˆ B

â‡’ AB < AC (Side opposite to the larger angle is larger.)

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