Question

In triangle ABC, AB=AC.The circle through B touches the side AC at the mid-point D of AC, passes through a point P on AB. Prove that 4 ×AP = AB ?

Answer 1

see explanation.

Explanation:

According to tangent-secant theorem:
"When a tangent and a secant are drawn from one single external point to a circle, square of the length of tangent segment must be equal to the product of lengths of whole secant segment and the exterior portion of secant segment."

`=> AD^2=AP*AB`
Given `AB=AC, and D` is midpoint of `AC`,
`=> AD=(AC)/2=(AB)/2`

`=>((AB)/2)^2=AP*AB`
`=> (AB)^2/4=AP*AB`
`=> AB=4AP`

Hence `4AP=AB`