Parallelism in circle using inscribed angles

Question

Let $[AB]$ be a chord in a circle.

$M$ is a point on the extension $[AB)$ outside the circle.

One tangent to the circle through $M$ meets the circle at $E$. $H$ and $K$ (resp.) are orthogonal projections of points $E$ and $B$ on lines $(AB)$ and $(ME)$.

it is asked to prove that $(HK) \parallel (AE)$

Any ideas are welcome.

thanks

Answer 1

Due to the right angles at $H$ and $K$, the quadrilateral $BEKH$ is cyclic, so we have:

$$ME\cdot MK=MB\cdot MH$$

Also, as $ME$ is a tangent, we have:

$$MA\cdot MB=ME^2$$

From those equalities, one easily derives (by multiplying them together):

$$MA\cdot MK=ME\cdot MH$$

which means that $\frac{MH}{MA}=\frac{MK}{ME}$, which makes triangles $MHK$ and $MAE$ similar - and then $HK\parallel AE$ easily follows.