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.

## Comments

Sort by:

TopNewestSimply 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 \(PQ || BC. \square\) – Mathh Mathh · 2 years, 3 months ago

Log in to reply