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 6 of 6: Conclusion: tangency at H
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AKKH=ALLH  ⟹  BD tangent to ⊙(SHT) at H\frac{AK}{KH}=\frac{AL}{LH} \implies BD \text{ tangent to } \odot(SHT) \text{ at } H
Detailed analysis

The ratio AKKH=ALLH\dfrac{AK}{KH}=\dfrac{AL}{LH} established above is exactly the condition (via the angle-bisector length theorem) that makes the bisectors of ∠AKH\angle AKH and ∠ALH\angle ALH meet on segment AHAH, which is equivalent to line BDBD (perpendicular to AHAH at HH) being tangent to the circumcircle of triangle SHTSHT at HH, as reduced earlier. This completes the proof.