Question

If the bisector of the vertical /_Bac of a triangle ABC, is perpendicular to the base BC, prove that the triangle is an isosceles triangle.​

Answer 1

Step-by-step explanation:

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=2180