Question

given: ae ≅ ce ; de ≅ be
prove: abcd is a parallelogram.
we have that ab || dc.
by a similar argument used to prove that △aeb ≅ △ced, we can show that △ ≅ △ceb by
. so, ∠cad ≅ ∠ by cpctc.
therefore, ad || bc by the converse of the
theorem. since both pair of opposite sides are parallel, quadrilateral abcd is a parallelogram.

Explanation:

Step1: Identify congruent triangles

In $\triangle AED$ and $\triangle CEB$, we have $AE\cong CE$, $DE\cong BE$ and $\angle AED=\angle CEB$ (vertically - opposite angles). So, $\triangle AED\cong\triangle CEB$ by the Side - Angle - Side (SAS) congruence criterion.

Step2: Use CPCTC

By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), $\angle CAD\cong\angle ACB$.

Step3: Apply converse of alternate - interior angles theorem

Since $\angle CAD\cong\angle ACB$, by the converse of the alternate - interior angles theorem, $AD\parallel BC$. We already know that $AB\parallel DC$. Since both pairs of opposite sides are parallel, quadrilateral $ABCD$ is a parallelogram.

Answer:

In the blanks: $\triangle AED$, SAS, $\angle ACB$, alternate - interior angles.