Problem 3

Convex quadrilateral

ABCD

has

\angle ABC=\angle CDA=90^\circ

. Point

H

is the foot of the perpendicular from

A

BD

. Points

S

and

T

lie on sides

AB

and

AD

, respectively, such that

H

lies inside triangle

SCT

and

\angle CHS-\angle CSB=90^\circ

\angle THC-\angle DTC=90^\circ

. Prove that line

BD

is tangent to the circumcircle of triangle

TSH

Step 3 of 6: Reduce tangency to a bisector-meeting condition

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BD \text{ tangent to } \odot(SHT) \iff \text{bisectors of } \angle AKH, \angle ALH \text{ meet on } AH

Detailed analysis

Since $KH=KS$ and $LH=LT$ (both being circumradii), $H$ and $S$ are reflections across the perpendicular bisector of $HS$ , which is the angle bisector of $\angle AKH$ inside the isosceles triangle $KHS$ ; similarly for $H,T$ and $\angle ALH$ . Line $BD$ is tangent to the circumcircle of $SHT$ at $H$ exactly when it is the reflection of line $AH$ appropriately, which reduces (by the angle-bisector characterization of the perpendicular bisectors of $HS$ and $HT$ ) to showing that these two bisectors meet on line $AH$ , and by the angle-bisector length theorem this is equivalent to $\dfrac{AK}{KH}=\dfrac{AL}{LH}$ .