The incircle of ΔABC touches the sides BC, CA and BA at D, E and F respectively. If
AB = AC, prove that BD = CD.
AF = AE
(Tangent from external point A)
BF = BD (Tangent from external point B)
and CD = CE (Tangent from external point C)
Now, AB = AC
⇒ AF + BF = AE + EC
⇒ BF = EC [∵ AF = AE]
⇒ BD = CD [ ∵ BF = BD; CD = CE] Proved.
A quadrilateral ABCD is drawn to circumscribe a circle (Fig. 10.62). Prove that AB + CD = AD + BC.