Question

A quadrilateral ABCD, AP and DP are the bisectors of angle A and angle B respectively. Show that Angle APD = 1/2 (angle B + angle C) and also draw the figure

Answer 1

So sum of angles of triangle = 180 

so from figure ∠APD + A/2 + D/2 = 180 ⇒ $\frac{A+D}{2}=180-P$

and sum of all angles of quadrilateral  = 360 

so A + B + C + D = 360 

⇒  $2\frac{A+D}{2}$ + B + C = 360 

⇒  2$(180-P)$ + B + C = 360 

⇒ 2P = B + C 

⇒  P  = 1/2 (B+ C) PROVED