Question

A circle circumscribes a ΔABC, DE is parallel to the tangent AP at A and intersects AB and AC in D and E respectively. Prove that:

1. ΔАВС ~ ΔAED 
2. AC × AE = AB × AD

Answer 1

Given:

A circle circumscribes ΔABC. DE is parallel to the tangent AP at A and intersects AB and AC in D and E, respectively.

To prove:

(i) ΔABC ~ ΔAED

Since DE ∥ AP,

∠ADE = ∠EAP (corresponding)

By the alternate segment theorem,

∠EAP = ∠ACB

∠ADE = ∠ACB

Since DE ∥ AP,

∠AED = ∠EAP (corresponding angles).

By the alternate segment theorem

∠EAP = ∠ABC.

Hence, ∠AED = ∠ABC.

Thus, by the AA similarity criterion,

ΔABC ∼ ΔAED

(ii) AC × AE = AB × AD

From the similarity of triangles, corresponding sides are proportional:

`(AB)/(AD) = (AC)/(AE)`

Cross multiply to get:

AC × AE = AB × AD