Advanced Geometric Problem & Solution (Twice a week update) Now 4563 problems posted.
<Qua-2024-732>, hint
Suppose AC and BD meet at P', then <AP'B=<CP'D=<BP'C=X
Then <AP'B+<BP'C+<CP'D+<DP'A=360
3X+180-X=360, X=90. Hence Triangles AP'B, BP'C, CP'D, DP'A are right
Hence 2MP'=AB, 2KP'=CD. P=P'
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