# D is a point on BC such that AD is the bisector of angleA. Show that: angle ADC = 90 + (angleB - angleC)/2

D is a point on BC such that AD is the bisector of angleA. Show that:
angle ADC = 90 + $\frac{\left(angleB-angleC\right)}{2}$
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losnonamern
e know that
= 180 - [180 - angle B - angle BAD]
= angle B + angle BAD
= angle B + angle A/2 since AD is an angular bisector
= angle B + (90 - angle B/2 -angle C/2)
= angle B + 90 - angle B/2 -angle C/2
= 90 + angle B/2 -angle C/2
= 90 + (angle B - angle C)/2thus proved