Question

1. given: rectangle abcd with e, the midpoint of \\(\overline{dc}\\). prove: \\(\angle eab \cong \angle eba\\) 43. given: parallelogram abcd and \\(\overline{ae} \cong \overline{cf}\\) prove: \\(\angle 1 \cong \angle 2\\)

Explanation:

Problem 42

Step1: Recall rectangle properties

In rectangle \(ABCD\), \(AD = BC\), \(\angle D = \angle C = 90^\circ\), \(AB = DC\).

Step2: Use midpoint definition

Since \(E\) is midpoint of \(\overline{DC}\), \(DE = EC = \frac{1}{2}DC\).

Step3: Prove \(\triangle ADE \cong \triangle BCE\)

By SAS congruence: \(AD = BC\), \(\angle D = \angle C\), \(DE = EC\), so \(\triangle ADE \cong \triangle BCE\).

Step4: Corresponding parts of congruent triangles

From congruence, \(\overline{AE} = \overline{BE}\).

Step5: Base angles of isosceles triangle

In \(\triangle EAB\), \(\overline{AE} = \overline{BE}\), so \(\angle EAB \cong \angle EBA\).

Step1: Recall parallelogram properties

In parallelogram \(ABCD\), \(AB = CD\), \(AB \parallel CD\), so \(\angle BAF = \angle DCE\).

Step2: Derive \(BF = DE\)

Given \(\overline{AF} \cong \overline{CF}\), and \(AC = AC\), so \(AC - AF = AC - CF\), i.e., \(\overline{BF} = \overline{DE}\).

Step3: Prove \(\triangle ABF \cong \triangle CDE\)

By SAS congruence: \(AB = CD\), \(\angle BAF = \angle DCE\), \(AF = CE\), so \(\triangle ABF \cong \triangle CDE\).

Step4: Corresponding parts of congruent triangles

From congruence, \(\angle ABF = \angle CDE\).

Step5: Use parallel line alternate angles

Since \(AB \parallel CD\), \(\angle ABD = \angle CDB\). Subtract: \(\angle ABD - \angle ABF = \angle CDB - \angle CDE\), so \(\angle 1 \cong \angle 2\).

Answer:

\(\angle EAB \cong \angle EBA\) is proven.

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Problem 43