Question

in the diagram below, $overline{ad}congoverline{ae}$, $overline{ba}paralleloverline{ce}$, $overline{cb}paralleloverline{da}$ and $mangle bad = 62^{circ}$. find $mangle dae$. you may assume lines that appear straight are straight, but the figure is not otherwise drawn to scale. answer attempt 1 out of 2 $mangle dae=square^{circ}$

Explanation:

Step1: Use parallel - line property

Since $BA\parallel CE$ and $\angle BAD = 62^{\circ}$, then $\angle ADC=\angle BAD = 62^{\circ}$ (alternate interior angles).

Step2: Consider the isosceles triangle

Given $AD\cong AE$, in $\triangle ADE$, let $\angle ADE=\angle AED$. And we know $\angle ADE = 62^{\circ}$ (from step 1).

Step3: Use angle - sum property of a triangle

In $\triangle ADE$, by the angle - sum property of a triangle $\angle DAE+ \angle ADE+\angle AED = 180^{\circ}$. Since $\angle ADE=\angle AED = 62^{\circ}$, then $\angle DAE=180^{\circ}-2\times62^{\circ}$.
$180 - 2\times62=180 - 124 = 56^{\circ}$.

Answer:

$56$