Advanced Geometric Problem & Solution (Twice a week update) Now 4603 problems posted.
<Tri-2025-2223>, hint
AH cuts BC in H'.
Since AD is the bisector of <A=90 , APDQ is a square and BD/CD=AB/AC
Triangles ABC,PBD, QDC are similar.
By Ceva's theorem (AP/PB)x(BH'/H'C)x(CQ/QA)=1
You can find AH' is altitude.
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