In the given figure, AC = AE.

Show that :

(i) CP = EP

(ii) BP = DP

(i) In ∆ADC and ∆ABE,

⇒ ∠ACD = ∠AEB [Angles in the same segment are equal]

⇒ AC = AE [Given]

⇒ ∠A = ∠A [Common]

Hence, ∆ADC ≅ ∆ABE by ASA axiom.

So, by C.P.C.T we have

⇒ AD = AB …………..(1)

Given,

⇒ AE = AC ………….(2)

Subtracting equation (1) from (2), we get :

⇒ AE - AD = AC - AB

⇒ DE = BC

In ∆BPC and ∆DPE,

⇒ ∠C = ∠E [Angles in the same segment are equal]

⇒ BC = DE [Proved above]

⇒ ∠CBP = ∠PDE [Angles in the same segment are equal]

Hence, ∆BPC ≅ ∆DPE by ASA axiom.

So, by C.P.C.T we have

⇒ CP = EP

Hence, proved that CP = EP.

(ii) Proved above,

∆BPC ≅ ∆DPE

∴ BP = DP [By C.P.C.T]

Hence, proved that BP = DP.