Center of circle tangent to hypotenuse in isosceles right triangle 
Let $ABC$ be a triangle with $AB=AC$ and $\angle BAC=90^{\circ}$. Let $P,Q$ be points on $AB, AC$ respectively such that $PQ=AB$, and let $O$ be the center of the circle passing through $P,Q$ and tangent to $BC$. Show that $OA\parallel BC$. 

It would be enough to show that $A,O,P,Q$ are concyclic. It seems like $\triangle OPQ\sim\triangle ABC$, which would imply the concyclic points, but I'm unsure how to do this. 
 A: Let the circle through points $P, \,Q$ and tangent to the segment $BC$ be denoted by $k$ and let the point of tangency of $k$ and $BC$ be denoted by $T$. Let line $PQ$ intersect line $BC$ at point $E$ and let us assume, without loss of generality, that point $E$ is located so that $B$ is between $E$ and $C$, while $P$ is between $E$ and $Q$. Then by the tangent-secant theorem (or whatever you call that statement) $$ET^2 = EP \cdot EQ$$ 
Let the line through point $P$ perpendicular to $AB$ and the line through point $W$ perpendicular to $AC$ intersect at point $R$. Then $APRQ$ is a rectangle and $PQ = AR = AB = AC$.
Denote by $T^*$ the orthogonal projection of $R$ onto the edge $BC$. 
Claim. Point $T$ coincides with point $T^*$, i.e. $T\equiv T^*$ meaning that the circle $k$ touches edge $BC$ at point $T^*$
Proof: Draw the circle $k_A$ centered at $A$ and of radius $AB = AC = AR$. Then $k_A$ passes through the points $B, \, R$ and $C$. If you denote say $\angle \, BAR = \alpha$, and observe that the quads $BPT^*R$ and $CQT^*R$ are cyclic, you can chase a bunch of angles and find that $$\angle \, PQT^* = \frac{\alpha}{2} = \angle \, PT^*B$$ (in fact, one can even show that $T^*$ is the incenter of triangle $PQR \, $) 
Edit. $\angle \, PQT^* = \frac{\alpha}{2} = \angle \, PT^*B$
Proof of edit: First, angle $\angle \, RAB = \alpha$ is central for circle $k_A$. Hence, $\angle \, RCB = \frac{1}{2} \, \angle \, RAB = \frac{1}{2} \alpha$. By cyclicity of $CQT^*R$ 
$$\angle \, RQT^* = \angle \, RCT^* = \angle \, RCB =  \frac{1}{2} \alpha$$ However, $APRQ$ is a rectangle, so $\angle \, RQP = \angle \, RAP = \angle \, RAB = \alpha$. Hence
$$\angle \, PQT^* = \angle \, RQP - \angle \, RQT^* = \alpha - \frac{1}{2} \alpha = \frac{1}{2} \alpha$$ 
Second, $AR = AB$ so $ABR$ is an isosceles triangle, so $$\angle \, ABR = \angle \, ARB = 90^{\circ} - \frac{1}{2} \alpha$$ and since $RP$ is orthogonal to $AB$, 
$$\angle \, PRB = 90^{\circ} - \angle \, ABR = \frac{1}{2} \alpha$$
By cyclicity of $BPT^*R$ 
$$\angle \, PT^*B = \angle \, PRB = \frac{1}{2} \alpha$$ Thus $\angle \, PQT^* = \frac{\alpha}{2} = \angle \, PT^*B$.
The latter equality of angles yields, for example, that triangles $EPT^*$ and $ET^*Q$ are similar. Therefore, $$\frac{EP}{ET^*} = \frac{ET^*}{EQ}$$ which is equivalent to $$ET^{*2} = EP \cdot EQ = ET^2$$ meaning that $ET^* = ET$. Since, by construction, both points $T$ and $T^*$ lie on the line $BC$ and are on the same side of point $B$, one can conclude that $T^* \equiv T$.
Concluding the proof. Let  Let $M = PQ \cap AR$. Then $M$ is the midpoint of both diagonal segments $PQ$ and $AR$ of rectangle $APRQ$. Since the point of tangency $T$ of the circle $k$ and the segment $BC$ is the orthogonal projection of point $R$, the center of $k$ lies on the intersection of the line $RT$ and the orthogonal bisector of diagonal $PQ$ through $M$. Again, if you chase some angles, you get  $$\angle \, ORM = \angle \, TRA = 45^{\circ} - \alpha$$ and  $$\angle \, AMO = 90^{\circ} - \angle \, AMQ = 90^{\circ} - 2\,\alpha$$ Therefore, $$\angle \, ROM = \angle \, AMO - \angle \, ORM = 90^{\circ}- 2\,\alpha - 45 + \alpha = 45^{\circ} - \alpha = \angle \, ORM$$ which means that triangle $ORM$ is isosceles with $MO = MR$ which means that $$MO = MR = MA$$ Consequently, triangle $AOR$ is right-angled with $\angle \, AOR = 90^{\circ}$ so line $AO$ is orthogonal to line $RT$. However, $BC$ is also orthogonal to $BC$ (the claim). Thus $AO$ is parallel to $BC$. 
A: This isn't a proof, because I'm starting with the assumption $OA \parallel BC$. Per questioner's request, writing this up as an answer. 
Note that $O$ can't be equal to $A$ because $OA$ wouldn't be a line segment. 
Let $D$ be the midpoint of $BC$. The circle won't be tangent to $BC$ at $D$. Let's say it is tangent at $E$. Then $OE$ = $AD$ and $OE \parallel AD$. $AD = \sqrt{2}/2 * AB = OE$. Thus radius is $r=\sqrt{2}/2 * AB$. Since $P$ and $Q$ lie on circle $OP = OQ = r$. And it's given that $PQ = AB$. So $\triangle OPQ$ has two sides equal to $r$ and one side equal to $\sqrt{2}r$. 
Therefore it is a right triangle (by Pythagoras) with equal sides, so $\,\triangle OPQ \sim \triangle ABC$. Further $\triangle OPQ = \triangle ABD = \triangle ACD$.
