Question

given: abcd is a parallelogram. diagonals ac, bd intersect at e. prove: ae ≅ ce and be ≅ de. assemble the proof by dragging tiles to the statements and reasons columns.

Explanation:

Step1: Recall properties of parallelogram

In parallelogram \(ABCD\), \(AB\parallel CD\) and \(AB = CD\) (opposite - sides of a parallelogram are parallel and equal).

Step2: Identify alternate - interior angles

Since \(AB\parallel CD\), \(\angle BAE=\angle DCE\) and \(\angle ABE=\angle CDE\) (alternate - interior angles).

Step3: Prove triangle congruence

In \(\triangle ABE\) and \(\triangle CDE\), we have \(\angle BAE=\angle DCE\), \(AB = CD\), and \(\angle ABE=\angle CDE\). So, \(\triangle ABE\cong\triangle CDE\) by the Angle - Side - Angle (ASA) congruence criterion.

Step4: Use congruent - triangle properties

Since \(\triangle ABE\cong\triangle CDE\), then \(AE = CE\) and \(BE = DE\) (corresponding parts of congruent triangles are congruent).

Answer:

Statements	Reasons
\(\angle BAE=\angle DCE\), \(\angle ABE=\angle CDE\)	Alternate - interior angles
\(\triangle ABE\cong\triangle CDE\)	ASA congruence criterion
\(AE\cong CE\), \(BE\cong DE\)	Corresponding parts of congruent triangles are congruent