Geometry : Prove that $PE=PC$ Let $l$ be a line not intersecting circle $\omega$ that has center $O$. Draw $OP$ perpendicular to $l$ at point $P$ and draw $PA$ tangent to $\omega$ at point $A$. Extend $OA$ to cut $\omega$ again at point $B$ and cut $l$ at point $C$. $PB$ cuts $\omega$ at point $D$ and $AD$ cuts $l$ at point $E$. Prove that $PE=PC$.
My thought :
Pole of point $P$ passes point $A$.
Since $P, C, E$ are collinear, so polar of $P, C, E$ are concurrent.

From the Menelaus Theorem on $\triangle CBP$ and the line $A-D-E$ we have:
$$\frac{CA}{AB} \times \frac{BD}{DP} \times \frac{PE}{CE} = 1$$
So from this it's enough to prove that $\frac{CA}{AB} \times \frac{BD}{DP} = 2$. 
Now we have that $AB = 2R$, while from the power of point $P$ we get: $DP = \frac{PA^2}{PB}$. Using some well-known formulas for altitudes in right-angled triangles we have:
$$\frac{CA}{AB} \times \frac{BD}{DP} = \frac{CA \cdot BD \cdot PB}{2R \cdot AP^2} = \frac{BD \cdot PB}{2R^2} = \frac{AB^2}{2R^2} = \frac{4R^2}{2R^2} = 2$$
Hence the proof.
