# Class 9 NCERT Solutions- Chapter 7 Triangles – Exercise 7.3

### Question 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 Figure). 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.

Solution:

Given: âˆ†ABC and âˆ†DCB are isosceles âˆ†on the same base BC.

To show:

• Î”ABD â‰… Î”ACD
• Î”ABP â‰… Î”ACP
• AP bisects âˆ A as well as âˆ D.
• AP is the perpendicular bisector of BC.

i) in âˆ†ABD and âˆ†ACB

AB=AC

BD=CD

AD=AD

âˆ†ABDâ‰…âˆ†ACD ————-(S.S.S)

ii) in âˆ†ABP and âˆ†ACP

AB=AC

âˆ  BAPâ‰…âˆ CAP    [âˆ†ABDâ‰…âˆ†ACD   BY C.P.CT]

AP=AP   ———[common]

âˆ´[âˆ†ABDâ‰…âˆ†ACD         ———–[S.A.S]

iii) [âˆ†ABDâ‰…âˆ†ACD         ———–[S.A.S]

âˆ BAD=âˆ CAD

AD, bisects âˆ A

AP, bisects âˆ A   —————–1

In âˆ† BDP and âˆ†DPB

BD=CD   —————(GIVEN)

DP=PC     ———-[âˆ†ABâ‰… âˆ†ACP   C.P.C.T]

DP=DP         ———–[common]

âˆ´âˆ†BDPâ‰…âˆ†CDP    (S.S.S)

âˆ BDP=âˆ CDP  (C.P.C.T)

DP bisects âˆ D

AP bisects âˆ D  ——————-2

From 1 and 2, AP bisects âˆ  A as well as âˆ  D

iv) âˆ  AP +âˆ APC =180Â°        ————[linear pair]

âˆ APB=âˆ APC    ————-[âˆ†ABPâ‰…âˆ†ACP       C.P.CT]

âˆ APB + âˆ APC=180Â°

2 âˆ  APB=180Â°

âˆ APB=180/2=90Â°

BP=PC            (FROM ii)

âˆ´AP is âŠ¥ bisects of BC.

### (i) AD bisects BC                      (ii) AD bisects âˆ A.

Solution:

Given: AB=AC,  AD altitude

To Show:

(i) AD bisects BC                      (ii) AD bisects âˆ A.

In âˆ†ADB and âˆ†ADC

âˆ ADB=âˆ ADC   ——– ———–[each 90Â°]   R

AB=AC               ——————–[given]S

AD=AD             ——–[common]S

âˆ´ âˆ†ADB â‰…âˆ†ADC

BD=DC             ————-[c.p.c.t]

âˆ´AD bisects BC

âˆ 1=âˆ 2               ————-[c.p.c.t]

âˆ´AD bisects âˆ A

### Question 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‰… PQR

Solution:

Given:

AB=PQ

BC=QR

AM=PN

AM and PN are medians

To show:(i) Î”ABM â‰… Î”PQN           (ii) Î”ABC â‰… PQR

Solution: In  Î”ABM and Î”PQN

AB=PQ

AM=PN

because AM and PN are medians BC=QR

therefore   =1/2BC=1/2QR

âˆ´BM=QN

âˆ´) Î”ABM â‰… Î”PQN                    ———[S.S.S]

âˆ B=âˆ Q             ——–[c.p.c.t]

ii)now in       Î”ABC and  Î”PQR

AB=PQ                ———-[given]

âˆ B=âˆ Q             from (i)

BC=QR               —————-[given]

âˆ´ Î”ABC â‰… PQR                 [S.A.S]

### Question 4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Solution:

Given: altitude BE and CF are equal

To prove: Î”ABC is an isosceles  Î”

In Î”BEC and Î”CEB

âˆ E=âˆ F        —————-[each 90Â°] R

BC=BC       —————–[common]  H

BF=CF       —————-[given]  S

# Î”BEC â‰… Î”CEB     [R.H.S]

âˆ C=âˆ B        ————-[C.P.C.T]

In Î”ABC,

âˆ C=âˆ B

### Question 5. ABC is an isosceles triangle with AB = AC. Draw AP âŠ¥ BC to show that âˆ B = âˆ C.

Solution:

Given:

In âˆ†ABC,

AB=BC

AP âŠ¥ BC

to show that: âˆ B = âˆ C.

Solution:

In âˆ†APB and âˆ†APC

âˆ APB = âˆ APC         —————[ each 90Â°]  R

AB=AC               ——————-[given]      H

AP=AP         ——————–[common]     S

âˆ´âˆ†APB â‰… âˆ†APC            ———-[R.H.S]

âˆ B = âˆ C  —————[C.P.C.T]

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