Find AC

Since $\angle BAE = \angle BCD$ and $\frac{CD}{AB} = \frac{2}{3} = \frac{CB}{AE}$ we see that $\triangle BCD \sim \triangle EAB$ . In particular, we see $\angle DBC = \angle BEA$ .

Then $m\angle ABC = m\angle ABE + m\angle DBC = m\angle ABE + m\angle BEA = 180^\circ - m\angle BAE = 60^\circ$

Also, since $\triangle ABC$ is isosceles, we can see $\angle BAC = \angle BCA$ , and as we have just shown $m\angle ABC = 60^\circ$ , we can conclude that $\triangle ABC$ is equiangular, hence equilateral. That is, $AC = AB = \boxed{6}$

Triangles $\Delta ABC$ and $\Delta ACF$ are equilateral triangles of sides $6$ .
Triangles $\Delta ABE$ and $\Delta DFE$ are similiar.

First, we see that BAE and BCD each are 120 degrees and they are opposite to each other. So ABCD is a parallelogram => AB // CD => ABE = BDC => ABC = ABE + CBD = BDC + CBD = 180 degrees - BCD = 180 - 120
=> ABC = 60
We have the law of cosine: a^2 = b^2 + c^2 - 2.b.c.CosA Apply this law for ABC triangle, we have: AC^2 = AB^2 + BC^2 -2.AB.BC.CosABC = 6^2 + 6^2 - 2.6.6.Cos60 = 36 + 36 - 36 = 36 => AC = 6 Sorry if I’m not good at English

Since; Angle(BAC) = Angle(BCD) & BD||BE both are similar triangles. There BD:DE=2/3 or BD = 2/3(DE) & BE= DE + (2/3)DE= 5/3(DE). Let, Extend CD such that it meets AE at F. Since CD||AB (similar triangles); therefore DE||AB and [DEF]~[ABE] and m(DE)=2; so ABCF is a ||gm, so angle ABC = 60 Deg, therefore AC= 6