Question

unit 2 lesson 7 practice problems 1. prove that the parallelogram abcd has two congruent triangles, abd and cdb. 2. assume that c is the mid - point of ae and the mid - point of bd. prove that △acb≅△ecd.

Explanation:

Step1: Recall parallelogram properties

In parallelogram \(ABCD\), \(AB = CD\) and \(AD=BC\) (opposite - sides of a parallelogram are equal), and \(BD = BD\) (common side).

Step2: Apply SSS congruence criterion

By the Side - Side - Side (SSS) congruence criterion, in \(\triangle ABD\) and \(\triangle CDB\), since \(AB = CD\), \(AD = BC\), and \(BD=BD\), we have \(\triangle ABD\cong\triangle CDB\).

Step3: For the second problem, use mid - point properties

Since \(C\) is the mid - point of \(AE\), \(AC = EC\). Since \(C\) is the mid - point of \(BD\), \(BC = DC\). Also, \(\angle ACB=\angle ECD\) (vertically opposite angles are equal).

Step4: Apply SAS congruence criterion

By the Side - Angle - Side (SAS) congruence criterion, in \(\triangle ACB\) and \(\triangle ECD\), since \(AC = EC\), \(\angle ACB=\angle ECD\), and \(BC = DC\), we have \(\triangle ACB\cong\triangle ECD\).

Answer:

1. In parallelogram \(ABCD\), \(\triangle ABD\cong\triangle CDB\) by SSS congruence criterion.
2. Given \(C\) is the mid - point of \(AE\) and \(BD\), \(\triangle ACB\cong\triangle ECD\) by SAS congruence criterion.