1. ## circle/ iscosceles triangle

In the given fig., ABC is an isosceles triangle in which AB = AC. A circle through B touches the side AC at D and it intersects the side AB at P. If D is the mid-point of AC, prove AB = 4AP.

2. Hello, snigdha!

I used a (sort of) well-known theorem:

If a tangent and a secant are drawn to a circle from an external point,
the tangent is the mean proportional of the secant and the external segment of the secant.

$\displaystyle ABC$ is an isosceles triangle wherte $\displaystyle AB = AC.$
A circle through $\displaystyle B$ touches side $\displaystyle AC$ at $\displaystyle D$, and intersects side $\displaystyle AB$ at $\displaystyle P$.
If $\displaystyle D$ is the midpoint of $\displaystyle AC$, prove that: .$\displaystyle AB \,=\, 4\!\cdot\!AP$
Code:
              B
♥ o o
o  *  *     o
o   *     *     o
o   *        *    o
*           *
o  *              * o
o *                 ♥ P
o*                  o *
*                       *
* o                 o      *
*   o               o         *
*      o           o             *
♥   *   *   o ♥ o   *   *   *   *   ♥ A
D

We are given: .$\displaystyle AB \,=\,AC$

$\displaystyle D$ is the midpoint of $\displaystyle AC\!:\;\;AD \,=\,\tfrac{1}{2}AB \quad\Rightarrow\quad AD^2 \,=\,\tfrac{1}{4}AB^2$ .[1]

From the theorem, we have: .$\displaystyle AD^2 \:=\:AB\cdot AP$ .[2]

Substitute [1] into [2]: .$\displaystyle \tfrac{1}{4}AB^2 \:=\:AP\cdot AB \quad\Rightarrow\quad AB^2 \:=\:4\!\cdot\!AP\!\cdot\!AB$

. . Therefore: .$\displaystyle AB \:=\:4\!\cdot\!AP$