Question

1. given: $overline{ab}perpoverline{bc}$, $overline{bc}perpoverline{dc}$, $overline{db}$ bisects $angle abc$, $overline{ac}$ bisects $angle dcb$, $eb = ec$. prove: $\triangle beacong\triangle ced$. 4. given: $angle1congangle2$, $angle3congangle4$. prove: $overline{ac}congoverline{bd}$

Explanation:

Step1: Analyze given right - angles

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

Step2: Use angle - bisector properties

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

Step3: Consider equal sides

Given $EB = EC$.

Step4: Prove vertical angles equal

$\angle AEB$ and $\angle DEC$ are vertical angles, so $\angle AEB=\angle DEC$.

Step5: Apply ASA congruence

In $\triangle BEA$ and $\triangle CED$, we have $\angle ABD=\angle ACD$, $EB = EC$, $\angle AEB=\angle DEC$. By the Angle - Side - Angle (ASA) congruence criterion, $\triangle BEA\cong\triangle CED$.

for second problem:

Step1: Consider the two triangles

In $\triangle ABC$ and $\triangle DCB$.

Step2: Use given angle equalities

We know that $\angle 1=\angle 2$ and $\angle 3=\angle 4$, and $BC = CB$ (common side).

Step3: Apply ASA congruence

By the Angle - Side - Angle (ASA) congruence criterion, $\triangle ABC\cong\triangle DCB$.

Step4: Use congruent - triangle properties

Since corresponding parts of congruent triangles are congruent, if $\triangle ABC\cong\triangle DCB$, then $\overline{AC}\cong\overline{BD}$.

Answer:

$\triangle BEA\cong\triangle CED$ (by ASA)