The image above is given. D is found on side BC of the isosceles triangle ABC. G is the continuation of the line AB. Given: BC || FE. Given: $\frac{GF}{BF}$ = $\frac{AG}{AC}$ . Prove: AE $\perp$ BC.
My question is: Is the proportion necessary? Is it possible to just say that because $\triangle$ ABC is isosceles, then D is the midpoint of side BC etc.?
How can one solve this?
$\triangle FGE \sim \triangle BGC\\ \frac {GF}{BF} = \frac{GE}{EC}\\ \frac{GE}{EC} = \frac{AG}{AC}\\ \frac {EC}{AC} = \frac{GE}{AG}$
$\frac {EC}{AC} = \frac{GE}{AG} \implies AE$ bisects $\angle GAC$