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.
While I am fully capable of making a mistake, I don't believe that DAC
is ever twice angle GAC (except in the limit as D approaches infinity).
My thinking:
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.
But DAC = 60 degrees - ADC, and ADC is not 0 degrees (except in the extreme limit).
So DAC < 2 * (GAC)
'Dis Proof
