Question

1. triangle cda is the image of triangle abc after a 180° rotation around the mid - point of segment ac. triangle ecb is the image of triangle abc after a 180° rotation around the mid - point of segment bc.

a. explain why abcd and abec are parallelograms.

b. identify at least two pairs of congruent angles in the figure and explain how you know they are congruent.

Explanation:

Step1: Recall parallelogram property

A quadrilateral is a parallelogram if a pair of opposite sides are parallel and equal.

Step2: Analyze \(ABCD\)

Since \(\triangle CDA\) is the image of \(\triangle ABC\) after a \(180^{\circ}\) rotation around the mid - point of \(AC\), \(AB\) is parallel to \(CD\) and \(AB = CD\) (by the property of \(180^{\circ}\) rotation around the mid - point of a line segment). So \(ABCD\) is a parallelogram.

Step3: Analyze \(ABEC\)

Since \(\triangle ECB\) is the image of \(\triangle ABC\) after a \(180^{\circ}\) rotation around the mid - point of \(BC\), \(AB\) is parallel to \(CE\) and \(AB=CE\). So \(ABEC\) is a parallelogram.

Step4: Identify congruent angles in \(ABCD\)

In parallelogram \(ABCD\), \(\angle ABC\cong\angle CDA\) and \(\angle BAD\cong\angle BCD\) because opposite angles of a parallelogram are congruent.

Step5: Identify congruent angles in \(ABEC\)

In parallelogram \(ABEC\), \(\angle ABC\cong\angle CEB\) and \(\angle BAC\cong\angle BCE\) because opposite angles of a parallelogram are congruent.

Answer:

a. For \(ABCD\), \(\triangle CDA\) is the \(180^{\circ}\) rotation of \(\triangle ABC\) around the mid - point of \(AC\), so \(AB\parallel CD\) and \(AB = CD\), making \(ABCD\) a parallelogram. For \(ABEC\), \(\triangle ECB\) is the \(180^{\circ}\) rotation of \(\triangle ABC\) around the mid - point of \(BC\), so \(AB\parallel CE\) and \(AB = CE\), making \(ABEC\) a parallelogram.
b. In \(ABCD\): \(\angle ABC\cong\angle CDA\) (opposite angles of parallelogram \(ABCD\)); In \(ABEC\): \(\angle ABC\cong\angle CEB\) (opposite angles of parallelogram \(ABEC\)).