Question

In a triangle ABC, AB=AC and D is a point on side AC such that BD=BC. How to prove that BC×BC=AC×DC?

Answer 1

see explanation

Explanation:

Let `angleABC=x`,
given `AB=AC, => angleACB=x`,
let `angleBAC=y, => y+2x=180^@`
By the law of sines, we get :
`(BC)/siny=(AC)/sinx ------ Eq(1)`
Now, in `DeltaBCD`,
given that `BD=BC, => angleBDC=angleBCD=x`
`=> angleCBD=180-2x=y`
Similarly, by the law of sines, we get:
`(BC)/sinx=(DC)/siny ------ Eq(2)`
Multipying Eq(1) by Eq(2), we get:
`(BC)^2/(sinxsiny)=(AC)/(sinx)xx(DC)/siny`
`=> (BC)^2=AC*DC` ---(proved)