MathLabs

Problem 3

Convex quadrilateral ABCDABCD has ∠ABC=∠CDA=90∘\angle ABC=\angle CDA=90^\circ. Point HH is the foot of the perpendicular from AA to BDBD. Points SS and TT lie on sides ABAB and ADAD, respectively, such that HH lies inside triangle SCTSCT and ∠CHS−∠CSB=90∘\angle CHS-\angle CSB=90^\circ, ∠THC−∠DTC=90∘\angle THC-\angle DTC=90^\circ. Prove that line BDBD is tangent to the circumcircle of triangle TSHTSH.
Step 1 of 6: Locate the circumcentre K of SHC on line AB
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∠SQC=90∘−∠BSC=180∘−∠SHC  ⟹  C,H,S,Q concyclic\angle SQC = 90^\circ - \angle BSC = 180^\circ - \angle SHC \implies C,H,S,Q \text{ concyclic}
Detailed analysis

Let the line through CC perpendicular to SCSC meet line ABAB at QQ. From ∠CHS−∠CSB=90∘\angle CHS-\angle CSB=90^\circ one finds ∠SQC=90∘−∠BSC=180∘−∠SHC\angle SQC=90^\circ-\angle BSC=180^\circ-\angle SHC, so C,H,S,QC,H,S,Q lie on a common circle with diameter SQSQ; hence the circumcentre KK of triangle SHCSHC, being the midpoint of that diameter, lies on line ABAB.