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# Cool Geometry problem to prove

Suppose ABC is an arbitrary triangle with BC as its base. Extend BC in both ways such that XBCY straight line is created. Now bisect the two external angles ABX and ACY. From the vertex A drop two perpendicular lines on these bisectors - assume AP and AQ. Prove that PQ is parallel to BC.

Note by Raja Metronetizen
2 years, 12 months ago

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Simply extend $$AP$$ and $$AQ$$ (they meet the line $$BC$$ at $$P'$$ and $$Q'$$ respectively) and by $$ASA$$ you get equal triangles, hence $$AP=PP', AQ=QQ'$$, then $$PQ$$ is just the mid-segment of $$\triangle AP'Q'$$, which means $$PQ || BC. \square$$ · 2 years, 12 months ago