Question

A
1. AABC and A DBC are two isosceles triangles on
the same base BC and vertices A and D are on the
meside of BC (see Fig. 739) IfAD is extended
to intersec: BC at P, show that
A ABD =AACD
0
A ABP = AACP
AP bisects LA as well as ZD.
B
P
c
(iv) AP is the perpendicular bisector of BC.
Fig. 7.39​

Answer 1

Answer:

(i) In △ABD and △ACD,

AB=AC ....(since △ABC is isosceles)

AD=AD ....(common side)

BD=DC ....(since △BDC is isosceles)

ΔABD≅ΔACD .....SSS test of congruence,

∴∠BAD=∠CAD i.e. ∠BAP=∠PAC .....[c.a.c.t]......(i)

(ii) In △ABP and △ACP,

AB=AC ...(since △ABC is isosceles)

AP=AP ...(common side)

∠BAP=∠PAC ....from (i)

△ABP≅△ACP .... SAS test of congruence

∴BP=PC ...[c.s.c.t].....(ii)

∠APB=∠APC ....c.a.c.t.

(iii) Since △ABD≅△ACD

∠BAD=∠CAD ....from (i)

So, AD bisects ∠A

i.e. AP bisects∠A.....(iii)

In △BDP and △CDP,

DP=DP ...common side

BP=PC ...from (ii)

BD=CD ...(since △BDC is isosceles)

△BDP≅△CDP ....SSS test of congruence

∴∠BDP=∠CDP ....c.a.c.t.

∴ DP bisects∠D

So, AP bisects ∠D ....(iv)

From (iii) and (iv),

AP bisects ∠A as well as ∠D.

(iv) We know that

∠APB+∠APC=180°

....(angles in linear pair)

Also, ∠APB=∠APC ...from (ii)

∴∠APB=∠APC= 180/2 =90 °

BP=PC and ∠APB=∠APC=90 °

Hence, AP is perpendicular bisector of BC.