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