Isoceles Triangle Problem

Question

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Given

ABC is an Isosceles Triangle with $AB = AC$

$$DA = DC$$

Circle with Center $D$ and Radius $DA$ intersect $BC$ at $E$

To Prove

$$BE = CE$$

Answer 1

$E$ is the intersection of the circle centered at $D$ with radius $AD$, hence , a point on the circle. So the distance between $D$ and $E$ is exactly the radius of the circle. Thus, $DE=CD$ and $△DEC$ is also an isosceles triangle, and $∠ACB = ∠DCE$.

From the property of isosceles triangles we can say that $$ ∠ACB = ∠ABC,\ ∠DEC = ∠DCE, $$ which implies $$ ∠ABC = ∠DEC. $$ By AA post, $$△ABC \sim△DEC.$$

Note that$$ 2DC=AC \Longrightarrow \frac{BC}{EC} = \frac{AC}{DC} = 2 \Longrightarrow 2EC=BC, $$ so points $B, E, C$ are collinear points and $BC=BE+EC$. Therefore $BE=CE$.

I'm a student so I'm not that confident on my answer but I hope it can help.