Question

In the given figure, AC = AE. Show that:

1. CP = EP
2. BP = DP

Answer 1

In ΔADC and ΔABE,

∠ACD = ∠AEB  ...(Angles in the same segment)

AC = AE  ...(Given)

∠A = ∠A  ...(Common)

∴ ΔADC ≅ ΔABE  ...(ASA postulate)

${\displaystyle \Rightarrow }$ AB = AD

But AC = AE

∴ AC – AB = AE – AD

${\displaystyle \Rightarrow }$ BC = DE

In ΔBPC and ΔDPE

∠C = ∠E  ...(Angles in the same segment)

BC = DE

∠CBP = ∠CDE  ...(Angles in the same segment)

∴ ΔBPC ≅ ΔDPE   ...(ASA Postulate)

${\displaystyle \Rightarrow }$ BP = DP and CP = PE  ...(C.P.C.T.)