An easy problem

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It is given a circle k. From the point P are drawn the tangents PA and PB to the k. On the smaller arc AB is chosen a point Q. PQ intersects k at the point C. From Q is drawn a parallel line to AP that intersects AC at the point D. If M is the intersection point of AB and QD prove that M is the middle of QD

[Dimoniada]

На $AB$ лежит симедиана тр-ика $QAC$. Далее очевидно.

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Can you be more detailed if you have time?

[Dimoniada]

Let $T=QC\cap AB$. From similarity $\triangle APC \sim \triangle QPA$ and $\triangle BPC \sim \triangle QPB$ we obtain: $\frac {AC} {AQ} = \frac {PC} {AP}$, $\frac {BC} {BQ} = \frac {PC} {BP}$, i.e. $\frac {AC} {AQ} =\frac {BC} {BQ}$ ($AP=BP$). Next step $\triangle AQT \sim \triangle CBT$, $\triangle ATC \sim \triangle TQB$, so $\frac {BC} {CT} = \frac {AQ} {AT}$, $\frac {BQ} {QT} = \frac {AC} {AT}$ => $\frac {CT} {TQ} = \frac {AC^2} {AQ^2}$ and consequently $AT$ - symedian in $\triangle ACQ$. Finally $\triangle AQC \sim \triangle ADQ$ => $AM$ - median in $\triangle AQD$.

[ins-]

Thank you for the excellent solution. I (re)discovered the statement at my own. Hope you like it.