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 4 of 6: A midpoint M and the circumcentre O of ABCD

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M := KL \cap HC \implies M \text{ midpoint of } HC, \quad OM \parallel AH \perp BD

Detailed analysis

Let $M=KL\cap HC$ . Since $KH=KC$ and $LH=LC$ , points $H$ and $C$ are symmetric about line $KL$ , so $M$ is the midpoint of $HC$ . Because $\angle ABC=\angle ADC=90^\circ$ , quadrilateral $ABCD$ is cyclic with circumcentre $O$ the midpoint of $AC$ ; then $OM\parallel AH$ (midline of triangle $AHC$ ), so $OM\perp BD$ , and since $OB=OD$ , line $OM$ is the perpendicular bisector of $BD$ , giving $BM=DM$ .