Question

1. given: (overline{ab}perpoverline{bc}), (overline{bc}perpoverline{dc}), (db) bisects (angle abc), (overline{ac}) bisects (angle dcb), (eb = ec). prove: (\triangle beacong\triangle ced)

Explanation:

Step1: Identify right - angles

Since $\overline{AB}\perp\overline{BC}$ and $\overline{BC}\perp\overline{DC}$, $\angle ABC=\angle DCB = 90^{\circ}$.

Step2: Use angle - bisector property

$DB$ bisects $\angle ABC$ and $AC$ bisects $\angle DCB$, so $\angle ABD=\angle DBC = 45^{\circ}$ and $\angle ACD=\angle ACB = 45^{\circ}$.

Step3: Consider side - side - angle (SSA is not valid, but here we use ASA)

We know that $EB = EC$. Also, $\angle EBA=\angle ECD$ (from angle - bisector results) and $\angle AEB=\angle DEC$ (vertically opposite angles).

Step4: Apply ASA congruence criterion

In $\triangle BEA$ and $\triangle CED$, we have $\angle EBA=\angle ECD$, $EB = EC$, and $\angle AEB=\angle DEC$. So, by the Angle - Side - Angle (ASA) congruence criterion, $\triangle BEA\cong\triangle CED$.

Answer:

$\triangle BEA\cong\triangle CED$ by ASA congruence criterion.