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 5 of 6: Two circles through M give the sine-rule ratio
Previous stepNext step
AKAL=sin⁡∠ALKsin⁡∠AKL=DMCL⋅CKBM=CKCL=KHLH\frac{AK}{AL}=\frac{\sin\angle ALK}{\sin\angle AKL}=\frac{DM}{CL}\cdot\frac{CK}{BM}=\frac{CK}{CL}=\frac{KH}{LH}
Detailed analysis

Since CM⊥KLCM\perp KL, points B,C,M,KB,C,M,K lie on a circle with diameter KCKC, and similarly L,C,M,DL,C,M,D lie on a circle with diameter LCLC. Applying the sine rule in triangle AKLAKL and in these two circles (using BM=DMBM=DM from the previous step) gives AKAL=sin⁡∠ALKsin⁡∠AKL=DMCL⋅CKBM=CKCL=KHLH\dfrac{AK}{AL}=\dfrac{\sin\angle ALK}{\sin\angle AKL}=\dfrac{DM}{CL}\cdot\dfrac{CK}{BM}=\dfrac{CK}{CL}=\dfrac{KH}{LH}, which rearranges to exactly AKKH=ALLH\dfrac{AK}{KH}=\dfrac{AL}{LH}.