In fact, the problem is that d is a point on the ABC base of an obtuse isosceles triangle, DE⊥AB, DF⊥AC,

So we get DE+DF=CG. It just complicates the graphics.

Reason: sδ ABC = 1/2 * AB * CG,

Connect AD, sδ Abd = 1/2 * AB * DE, sδ ACD = 1/2 * AC * DF,

∫sδABC = sδABD+sδACD，

∴ 1/2*ab*cg= 1/2*ab*de+ 1/2*ac*df,∵ab=ac

∴CG=DE+DF。