Question

proving the converse of the parallelogram diagonal theorem
given: ( overline{ae} cong overline{ce}, overline{de} cong overline{be} )
prove: ( abcd ) is a parallelogram.
we have that ( overline{ab} parallel overline{dc} ).
by a similar argument used to prove that ( \triangle aeb cong \triangle ced ), we can show that ( \triangle ) ( cong \triangle ceb ) by ( ). so, ( angle cad cong angle ) ( ) by cpctc. therefore, ( overline{ad} parallel overline{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

We know that \(AE = CE\), \(DE = BE\), and \(\angle AED=\angle BEC\) (vertically - opposite angles). So, \(\triangle AED\cong\triangle CEB\) by \(SAS\) (Side - Angle - Side) congruence criterion.

Step2: Use CPCTC

Since \(\triangle AED\cong\triangle CEB\), by CPCTC (Corresponding Parts of Congruent Triangles are Congruent), \(\angle CAD=\angle ACB\)

Step3: Use converse of alternate - interior angles theorem

If \(\angle CAD=\angle ACB\), then \(AD\parallel BC\) by the converse of the alternate - interior angles theorem.

Answer:

The first blank is \(\triangle AED\), the second blank is \(SAS\), the third blank is \(\angle ACB\), and the fourth blank is alternate - interior angles.