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.

#Geometry #GeometricMeasurements

Note by Sayan Chaudhuri
6 years, 10 months ago

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Comments

Simply extend APAP and AQAQ (they meet the line BCBC at PP' and QQ' respectively) and by ASAASA you get equal triangles, hence AP=PP,AQ=QQAP=PP', AQ=QQ', then PQPQ is just the mid-segment of APQ\triangle AP'Q', which means PQBC.PQ || BC. \square

mathh mathh - 6 years, 10 months ago
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