Proving a line perpendicular to another in a circle

Question

I got a question at an exam recently, and was unable to solve it.

Let $\Gamma_1$ be a circle with a chord $CD$ and a diameter $AB\perp CD$ at $N$, with $AN\geq BN$. A circle $\Gamma_2$ is drawn with centre $C$ and radius $CN$. It intersects $\Gamma_1$ at $P,Q$. $PQ$ intersects $CD$ at $M$ and $AC$ at $K$. $NK$ extended meets $\Gamma_2$ at $L$. Prove that $AL\perp PQ$.

Here's the diagram:

I tried various approaches, like proving $AL$ tangent to $\Gamma_2$, Proving $OC||AL$, proving $QP||BF$, proving $QB=PF$, proving $CNAL$ cyclic, all to no avail.

Please help.

Answer 1

Proof

Let $CD$ intersect $\Gamma_2$ at a second point $S$. Notice that $PQ$ is the radical axis of $\Gamma_1$ and $\Gamma_2$. Thus $$(\overline{CN}+\overline{CM})\cdot(\overline{CN}-\overline{CM}) =\overline{SM}\cdot \overline{MN}=\overline{CM}\cdot \overline{MD}=(\overline{CN}-\overline{MN})\cdot(\overline{CN}+\overline{MN}), $$which implies $$\overline{CM}=\overline{MN}.\tag1$$ Morover, since$$\overline{CK}\cdot\overline{KA}=\overline{NK}\cdot\overline{KL},$$ then $A,N,C,L$ are concyclic. Therefore $$\angle ALC=\angle BNC=90^o,$$ which shows that $AL$ is tangent to $\Gamma_2$. By the symmetry of the tangents $AL,AN$, we may obtain $$\overline{LK}=\overline{KN}.\tag2$$ By $(1),(2)$,we have $$PQ // LC\perp AL.$$

Answer 2

Diagram

The diagram shows, $AG \parallel DF$ so $C\hat{F}L = E\hat{A}G = \delta$ and $E\hat{M}F = E\hat{G}A = \gamma$. We can note that $Cl \perp AL$ and $\Delta CLF$ and $CL \perp AL$ then $\gamma + \delta = 90°$