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By bisector theorem $$\frac{AD}{CD}=\frac{AB}{BC}$$ But $BE=CD$ and $AB=AC$, thus $$\frac{AD}{BE}=\frac{AC}{BC}$$ Whic implies that $DE$ is parallel to $AB$, as desired

parmenides51 wrote: Let $ABC$ be an isosceles triangle with $AB = AC$. Point $D$ lies on side $AC$ such that $BD$ is the angle bisector of $\angle ABC$. Point $E$ lies on side $BC$ between $B$ and $C$ such that $BE = CD$. Prove that $DE$ is parallel to $AB$. Let $ BE = DC = b, AD = d, EC = c $, Then $ AB = b + d $, PO + or the bisector theorem we can propose: $$\frac{AD}{CD}=\frac{AB}{BC}$$, then $ b ^ 2 = dc $, on the other hand by thales' theorem $ AB // DE $ if and only if $$\frac{b}{d}=\frac{b}{c}$$, which is equal to $ b ^ 2 = dc $, therefore $ AB // DE $

Let angle ABC= angle ACB= 2x(AB= AC) so, angle ABD=angle DBE x Let angle DEC= y so, angle DEB= 180°-y so, angle BDE=y-x In ∆ BDE, sin x/ DE= sin (y-x) / BE BE/ DE= sin(y-x)/ sin x In ∆ EDC, sin y/ DC= sin 2x/ DE BE/ DE= sin y/ sin 2x( BE= CD) sin(y-x)/ sin x=sin y/ sin 2x 2 cos x sin(y-x)= sin y sin y+ sin ( y-2x)= sin y sin(y-2x)= 0 sin(y-2x)= sin 0 y-2x= 0 y= 2x So, angle DEC=y=2x=angle ABC since, both are corresponding angles, Therefore AB|| DE