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 re: Question? by Bractals)

Bractals

Here are the vertices:

• A = (0,0)
• B = (5/2,-2/3)
• C = (5/2,13/3)
• D = (0,5)
• E = (-5/2,13/3).

All triangles are congruent and all angles are convex:

• ABC = ACD = ADE =75 degrees
• BAC = ADC = AED = 75 degrees
• ACB = CAD = DAE = 30 degrees;

when ABCD is a parallelogram with opposite angles of 75 and 115 degrees, and ACDE is a kite with angles 60, 75, 75, and 150 degrees.

Edited on October 18, 2010, 2:57 pm

Posted by broll on 2010-10-18 03:53:25