Question 67
In a convex quadrilateral $ABCD$, $\angle ABC=\angle CDA=90^\circ$. The perpendicular from $A$ to $BD$ meets $BD$ at $H$. Points $S,T$ lie on sides $AB,AD$, respectively, such that $H$ is inside triangle $SCT$ and
$$\angle CHS-\angle CSB=90^\circ,\qquad\angle THC-\angle DTC=90^\circ.$$
Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.
An admissible symmetric configuration satisfying both angle conditions; the conclusion circle is not drawn.
$$\angle CHS-\angle CSB=90^\circ,\qquad\angle THC-\angle DTC=90^\circ.$$
Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.
An admissible symmetric configuration satisfying both angle conditions; the conclusion circle is not drawn.