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.
AP = AR
BP = BQ
CR = CQ
[Tangents from internal points are equal.]
Adding, we get
AP + BP + CQ = AR + BQ + CQ
⇒ (AP + BP) + CR = AR + (BQ + CQ)
⇒ AB + 5 = 7 + BC
⇒ 10 + 5 = 7 + BC
⇒ BC = 15 – 7 = 8 cm.