Let ABCDE be a convex pentagon such that the
triangles ABC, ACD, and ADE are similar.
The diagonals BD and CE intersect in point P.

Prove that the ray AP bisects side CD.

(In reply to Question? by broll)

 

Angle ABC = Angle ACD = Angle ADE

Angle BAC = Angle CAD = Angle DAE

Angle ACB = Angle ADC = Angle AED

I don't see how you can fit together triangles

ABC and ACD to form a parallelogram.

Angle BAD would be 60 degrees and

angle BCD would be 150 degrees.

Those two angles would have to be equal

for ABCD to form a parallelogram.

Posted by Bractals on 2010-10-17 23:57:48