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.

If we can show that CAG:CAD is constant it is sufficient to look at the limit at AD goes to infinity.

When this happens, DEC = EDA => 0
CAD => 60
FE => 1/2 BC
EFG => 60
GAC => 30

CAG:CAD = 2:1

Posted by Eric on 2006-11-09 21:18:59