The incircle of ΔABC touches the sides BC, CA and BA at D, E and F respectively. If
AB = AC, prove that BD = CD.
A quadrilateral ABCD is drawn to circumscribe a circle (Fig. 10.62). Prove that AB + CD = AD + BC.
Since, the tangents to a circle from an exterior point arc equal in length.
Length of tangents from same external point are equal
AN = AM, BN = BL
and CL = CM
On adding, we get:
AN + BN + CL =
AM + BL + CM
⇒ AN + BN + CL = (AM + CM) + BL
⇒ AB + CL = AC + BL
⇒ CL = BL
⇒ The point L bisects BC.