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In any triangle ABC, if the angle bisector of `/_A`and perpendicular bisector of BCintersect, prove that they intersect on the circumcircle of the triangle ABC |
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Answer» Angle bisector`angleA`meets the circumcircle of triangleABC at point P from diagram `angleBAD=angleCAD` so, ABD is a Cyclic Quadrilateral `angleDBC=angleDAC=1/2angleA` `angleDCB=angle BAD=1/2angleA` `angleDBC=angleDCB` `In triangleDCB` BP and CD are equal D lies on perpendicular bisector of BC |
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