ΔABC is equilateral. Point D lies on line BC such that C lies between B and D. Point E lies on side AC such that ED bisects angle ADC. Point F lies on side AB such that FE and BC are parallel. Point G lies on side BC such that GF=EF.

Prove that angle DAC equals twice angle GAC.

(In reply to 'Dis Proof by Steve Herman)

ABC is equilateral, FE is parallel to BC, and GF = EF.  This implies that G bisects BC, triangle GAC is a right triangle, and GAC = 30 degrees.

G does not bisect BC (unless E bisects AC which it doesn't unless D is a point at infinity.)

I've not solved this either but playing with Geometer's Sketchpad has convinced me it is true.

Posted by Jer on 2006-11-10 10:46:20