A circle with diameter $AC$ is intersected by a secant at points $B,D$. The secant and the diameter intersect at point $P$ outside the circle.

Question

A circle with diameter $AC$ is intersected by a secant at points $B,D$. The secant and the diameter intersect at point $P$ outside the circle. Perpendiculars $AE$ and $CF$ are drawn at the secant. If $EB = 2$ and $BD = 6$, find $DF$.

What I Tried: Here is a picture :-

Unfortunately I have no idea to start with this. A first idea is to find similar triangles, which I find a lot, but I cannot find a way on how to use the lengths of $EB$ and $BD$.

Can anyone help me? Thank You.

Answer 1

Denote the center of the circle as $O$. Draw a perpendicular line from $O$ to the secant, and denote the point of intersection $Q$.

We can prove that $BQ = QD$. Moreover $EA // QO // FC$.

Since $AO = OC$ as radii, $EQ = QF$ by intercept theorem.

Now $FD = QF - QD = EQ - BQ = EB = 2$.

Answer 2

$\angle ADE = \angle DCF\Rightarrow \triangle AED \sim \triangle DFC $ $$\Rightarrow \dfrac{AE}{DF} = \dfrac{AD}{DC} $$

$\angle ABE = \angle ACD\Rightarrow\triangle AEB \sim \triangle ADC $ $$\Rightarrow \dfrac{AE}{EB} = \dfrac{AD}{DC} $$

$$\therefore DF=EB =2$$

Answer 3

Let's call the point of intersection of line FC with circle as M. If we chase the angles we find BA=DM,AE=FM and angle BEA=DFM So, BAE is congruent to DMF and DF=EB=2