Question

in parallelogram (abcd), (mangle adc = 100) and (overline{ae}) bisects (angle dab). what kind of triangle must (\triangle ade) be?

Explanation:

Step1: Find the measure of \(\angle DAB\)

In a parallelogram \(ABCD\), \(\angle ADC+\angle DAB = 180^{\circ}\) (adjacent angles of a parallelogram are supplementary). Given \(m\angle ADC=100^{\circ}\), then \(m\angle DAB=180 - 100=80^{\circ}\).

Step2: Find the measure of \(\angle DAE\)

Since \(\overline{AE}\) bisects \(\angle DAB\), \(m\angle DAE=\frac{1}{2}m\angle DAB\). Substituting \(m\angle DAB = 80^{\circ}\), we get \(m\angle DAE = 40^{\circ}\).

Step3: Find the measure of \(\angle DEA\)

In parallelogram \(ABCD\), \(AB\parallel DC\), so \(\angle DEA=\angle EAB\) (alternate - interior angles). And since \(\angle DAE=\angle EAB\) (angle - bisector), \(\angle DAE=\angle DEA = 40^{\circ}\).

Answer:

A. isosceles