Question

question
given: \\( \overline { d e } \cong \overline { c e } , \overline { a d } \perp \overline { d b } \\) and \\( \overline { b c } \perp \overline { c a } \\).
prove: \\( \triangle a e b \\) is an isosceles triangle.

answer
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Explanation:

Step1: Prove $\triangle ADE\cong\triangle BCE$

Since $\overline{AD}\perp\overline{DB}$ and $\overline{BC}\perp\overline{CA}$, $\angle ADE = \angle BCE=90^{\circ}$.
We know that $\overline{DE}\cong\overline{CE}$ (given).
Also, $\angle AED=\angle BEC$ (vertically - opposite angles).
By the AAS (Angle - Angle - Side) congruence criterion, $\triangle ADE\cong\triangle BCE$.

Step2: Show $AE = BE$

From the congruence of $\triangle ADE$ and $\triangle BCE$, we can conclude that $AE = BE$ (corresponding parts of congruent triangles are congruent).

Answer:

Since $AE = BE$, by the definition of an isosceles triangle (a triangle with at least two congruent sides), $\triangle AEB$ is an isosceles triangle.