Inscribed and circumscribed circle have the same midpoint $"$Let $ABC$ be a triangle with $\sphericalangle ACB=90°$. Let $H$ be the intersection of the line segment $AB$ and the height $h$ with respect to $C.$ Furthermore, let $P$ and $Q$ be the intersection of the line segment $AB$ and the angle bisectors of the angles $\sphericalangle ACH$ and $\sphericalangle HCB$.
Show that the centre of the circumscribed circle of $PQC$ is the same as the centre of the inscribed circle of $ABC."$
I'm really helpless with this problem, appreciate all your help.
Just a quick sketch. $M$ is the centre of both circles:

Here are my thoughts so far (thanks for the comment pointing this out):
The way I see it, there are two different ways to approach this; either assume that the inscribed circle is given and try to show that it is concentric to the circumscribed circle of the inner triangle. This could be done either by showing that the perpendicular line segment bisector intersects in $M$ or that the distance between $M$ and $P$ is the same as between $M$ and $Q$ and $M$ and $C.$
You could also try the other way around.
My main problem is, that I am a rookie at elementary geometry and have no experience in proofs for that matter.
 A: Based on your picture, let's assume $M$ is the center of the circumscribed circle of $PCQ$.
Now consider the line segment $MC$.
We first show that $MC$ is the bisector. Notice that:
$$\angle HCA=90^{\circ}-\angle A \implies \angle PCA=\angle HCP=\angle 45^{\circ}-\angle \frac{A}{2}\implies \angle HPC= 45^{\circ}+\angle \frac{A}{2};$$
hence: $\angle QCM=90^{\circ}-\angle QPC=45^{\circ}-\angle \frac{A}{2}$ (because we assumed $M$ is the center of the circumscribed circle). As a result, we have: $\angle QCM=\angle PCA$.
Now, realize that $\angle QCP=45^{\circ}$. Therefore:
$$45^{\circ}= \angle QCP=\angle QCM +\angle MCP=\angle PCA +\angle MCP=\angle MCA\implies \angle MCA=45^{\circ}.$$
Since we assumed $M$ is the center of the circumscribed circle and $\angle QCP=45^{\circ}$, we easily conclude $\angle AQM=45^{\circ}$. By using the law of sine in $AQM$ and $AMC$, we get:
$$\frac {\sin \angle QAM}{\sin 45^{\circ}}=\frac {QM}{AM}=\frac {MC}{AM}=\frac {\sin \angle MAC}{\sin 45^{\circ}};$$
hence $\angle QAM=\angle MAC$ (simply because $\angle A<90^{\circ}$), which means $AM$ is the bisector as well.
