Question

the proof that $\triangle acb \cong \triangle ecd$ is shown. given: $\overline{ae}$ and $\overline{db}$ bisect each other at $c$. prove: $\triangle acb \cong \triangle ecd$ what is the missing statement in the proof? $\bigcirc$ $\angle bac \cong \angle dec$ $\bigcirc$ $\angle acd \cong \angle ecb$ $\bigcirc$ $\angle acb \cong \angle ecd$ $\bigcirc$ $\angle bca \cong \angle dca$

Explanation:

To prove \(\triangle ACB \cong \triangle ECD\), we start with the given that \(\overline{AE}\) and \(\overline{DB}\) bisect each other at \(C\). This means \(AC = EC\) and \(BC = DC\) (by the definition of bisecting segments). For triangle congruence, we can use SAS (Side - Angle - Side) criterion. The included angle between the two sides in \(\triangle ACB\) and \(\triangle ECD\) should be equal.

\(\angle ACB\) and \(\angle ECD\) are vertical angles. By the Vertical Angles Theorem, vertical angles are congruent. So \(\angle ACB\cong\angle ECD\).

Let's analyze the other options:

• \(\angle BAC\cong\angle DEC\): We don't know this from the given information (bisecting segments) and it's not a vertical angle or a directly deducible angle for the SAS criterion here.
• \(\angle ACD\cong\angle ECB\): These are also vertical angles, but they are not the included angles for the triangles \(\triangle ACB\) and \(\triangle ECD\) that we need for SAS.
• \(\angle BCA\cong\angle DCA\): There is no reason to assume these angles are congruent from the given that the segments bisect each other.

Answer:

\(\boldsymbol{\angle ACB \cong \angle ECD}\) (the third option: \(\angle ACB \cong \angle ECD\))