Question

In the diagram below, AD‾≅AE‾,AD ≅ AE , BA‾∥CE‾,BA ∥ CE , CB‾∥DA‾CB ∥ DA and m∠BAD=62∘.m∠BAD=62 ∘ . Find m∠DAE.m∠DAE.

Answer 1

Given that

$\bar{AD}\cong\bar{AE}$

Therefore,

$\begin{gathered} m\angle BAD\cong m\angle ADE=62^0 \\ m\angle BAD\cong m\angle AED=62^0 \\ m\angle DAE=? \end{gathered}$

Hence,

$\begin{gathered} m\angle ADE+m\angle AED+m\angle DAE=180^0 \\ 62^0+62^0+m\angle DAE=180^0 \\ 124^0+m\angle DAE=180^0 \\ m\angle DAE=180^0-124^0 \\ m\angle DAE=56^0 \end{gathered}$

Therefore,

$m\angle DAE=56^0$

Reason

Sum of angles in a triangle.