# Question #72b8f

Mar 3, 2017

$x = {20}^{\circ}$

#### Explanation:

As shown in Fig 1,
$\angle A C B = 180 - \left(80 + 80\right) = {20}^{\circ}$
$\angle A D B = 180 - \left(80 + 60\right) = {40}^{\circ}$
$\angle A E B = 180 - \left(70 + 80\right) = {30}^{\circ}$

As shown in Fig 2,
Draw a line $D F$, which is parallel to $A B$,
$\implies \angle E F D = \angle C D F = {80}^{\circ}$
$\implies \angle F D B = {60}^{\circ}$
Draw a line $A F$
let $G$ be the intersection point of $A F \mathmr{and} B D$
$\implies \angle B A G = 60 , \implies \angle A G B = 60$
$\implies \angle D G F = 60 , \implies \Delta D G F$ is equilateral.
$\implies F G = G D = F D$

As $\angle A C F = \angle C A F = {20}^{\circ} , \Delta C F A$ is isosceles.
$\implies C F = F A$

Draw a line $C G$, which bisects $\angle A C B$,
$\implies \Delta A C G \mathmr{and} \Delta C A E$ are congruent triangles. (they share one common side $A C$ and their 3 corresponding angles are equal).
$\implies A G = C E$
Since $C F = F A , \mathmr{and} C E = A G , \implies F E = F G$
Since $F G = F D , \implies F E = F D , \implies \Delta F E D$ is isosceles.
$\implies \angle F E D = x + 30 = \frac{1}{2} \left(180 - 80\right) = 50$
$\implies x = {20}^{\circ}$