#### Question

In the given figure, AC = AE Show that:

(i) CP = EP

(ii) BP = DP

#### Solution

In ΔADC and ΔABE ,

`∠`ACD = `∠`AEB (angles in the same segment)

AC = AE (Given)

`∠`A = `∠`A(common)

∴ Δ ADC ≅ Δ ABE (ASA postulate)

⇒ AB = AD

But AC = AE

∴ AC – AB = AE – AD

⇒ BC = DE

In Δ BPC and ΔDPE

`∠`C = `∠`E (angles in the same segment)

BC = DE

`∠`CBP = `∠`CDE (angles in the same segment)

∴ ΔBPC ≅ ΔDPE (ASA Postulate)

⇒ BP = DP and CP = PE (cpct)

Is there an error in this question or solution?

Solution In the Given Figure, Ac = Ae Show That: (I) Cp = Ep (Ii) Bp = Dp Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.