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.
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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
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