Question

if $\angle bce \cong \angle dce$ and $be = 10$, what is $de$? diagram: right angles at $b$ (on ray $b$) and $d$ (on ray $d$), points $c, d, e, b$ connected; $de = \square$; submit button

Explanation:

Step1: Identify congruent triangles

We know that $\angle BCE \cong \angle DCE$, $\angle CBE = \angle CDE = 90^\circ$ (right angles), and $CE$ is common to both $\triangle CBE$ and $\triangle CDE$. By the Angle - Angle - Side (AAS) congruence criterion, $\triangle CBE \cong \triangle CDE$.

Step2: Use congruent triangle property

In congruent triangles, corresponding sides are equal. Since $\triangle CBE \cong \triangle CDE$, the corresponding side of $BE$ in $\triangle CDE$ is $DE$. Given that $BE = 10$, we can conclude that $DE=BE$.

Answer:

$10$