A challenge geometrical Let $ABC$ a right isosceles triangle and $M$ the middle point at the hypotenuse $AC$.
Inside the triangle, draw a circle that is tangent to $AB$ at $P$ and to $BC$ at $Q$.The line  $MQ$ cuts newly to the circle in $T$.If $H$ is the orthocenter of the triangle $AMT$,prove that $MH=BQ$.
I think that we should prove that $MH$ is equal to the radius of the circle but maybe exist another idea.
 A: Let $O$ be the center of the circle. 
Claim: $OMQC$ is a cyclic quadrilateral.

 Proof: This follows because $\angle OMC = \angle OQC = 90^\circ$.

Claim: $MOAT$ is a cyclic quadrilateral.

 Proof: This follows because $\angle OTM = \angle OQM = \angle OCM = \angle OAM $.

Claim: $OAT$ and $OCQ$ are congruent triangles

 Proof: $OA=OC, OT=OQ$ and $\angle TOA = \angle TMA = \angle QMC = \angle QOC$.

Claim: As Blue stated, $AT$ is tangential to the circle.

 Proof: $AT = QC = AP$.

Claim: $MOTH$ is a parallelogram.

 Proof: $\angle TMH = 90^\circ - \angle ATM = \angle MTO $ and $ \angle HTM = 90^\circ - \angle TMA = \angle TMO $. Thus the opposite sides are parallel $_\square$

Hence $MH = TO = OP = BQ$.
A: Proof without words.





Okay, here are some words: Step 1 shows that $\square AMOP$ and $\square AMTP$ are cyclic (opposite angles are supplementary); since they share three vertices, their circumcircles coincide, and we conclude that $\angle ATO$, which subtends a semi-circle, is a right angle.
In Step 2, various perpendicularities (the only non-obvious one being handled by Step 1) imply parallelisms that make $\square MOTH$ a parallelogram.
