ω is a circle with diameter AB. Points C,D lie on ω such that C,D are on different sides of AB. A line passing through C and parallel to AD cuts AB at F, and a line passing through D and parallel to AC cuts AB at E. A,B,C and D are in a way that E,F are inside ω. The line passing through E and perpendicular to AB cuts BD at X and the line passing through F and perpendicular to AB cuts BC at Y. Prove that the perimeter of triangle △AXY equals to 2CD.
Solution
Let K be the intersection point of CD, AX and L be the intersection point of CD, AY.
Then, since DAEX is a cyclic quadrilateral, we have ∠KDX=∠BAC=∠AED=∠AXD⟹DK=KX=AK. Similarly, AL=LY=LC. By Thales's theorem, we have XY=2KL. Therefore AX+XY+AY=AK+KL+LA=KD+KL+LC=CD Which completes the proof.
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Source: MathNet,
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