Question

2. In TraingleABC, AngleB = angleC. If AD, the bisector of AngleA meets BC at D, prove that D is the mid point of BC and AD is
perpendicular to BC.

Please answer!!

Answer 1

Answer:

given:

angleA= angleB

therefore AB=AC( side opposite to equal angle)-©

now;

in ∆ABD and ∆ACD we have

AB= AC ( from©)

AD= AD (common side)

Angle ADB= angle ADC=90°

therefore ∆ABDCongurance ∆ACD

HENCE

BD= CD(CPCT)

THEREFORE D IS THE MID POINT