The same as $\angle FAC$ ?

By angle addition, $\angle BAC = \angle BAD + \angle DAF + \angle CAF$ , and since we are given that $\angle BAC = 100°$ and $\angle CAF = 20°$ , $\angle BAD + \angle DAF = 80°$ . We are also given that $\angle DAF = \angle BAD + \angle CAF$ , so $\angle DAF = \angle BAD + 20°$ . Solving these two formulas gives $\angle DAF = 50° and \angle BAD = 30°$ .

Since $AB = AC$ and $\angle DAF = \angle BAD + \angle CAF$ , we can make a folds on segment $AD$ and $AF$ such that $B'$ and $C'$ meet at the same point $G$ , as shown below.

Let $H$ be the intersection of $DF$ and $AG$ . Since $\angle ACF = \angle ABD$ (from isosceles $\triangle ABC$ ), $\angle AGF = \angle DGA$ , and by triangle ratios $\frac{DG}{FG}=\frac{DH}{HF}$ . Since $GD = BD$ and $FG = FC$ by reflection, $\frac{BD}{FC}=\frac{DH}{HF}$ . Since we are given that $\frac{BD}{FC}=\frac{DE}{EF}$ , $E$ is the same point as $H$ .

Therefore, since $\angle DAE = \angle BAD$ by reflection, $\angle DAE = \angle BAD = \boxed{30°}$ .

Call $\angle BAD=\theta$ so $\angle DAF=\theta + 20^{\circ}$ . Also because $\triangle ABC$ is isoscles, $\angle ABC = \angle ACB = 40^{\circ}$

Then we can find exterior angles $\angle ADE= 40^{\circ}+\theta$ and $\angle AFE = 60^{\circ}$ , so in $\triangle DAF$ , $\angle DAF = 80^{\circ}-\theta$ .

Equating the two formulas for $\angle DAF$ gives $\theta=30^{\circ}$ and also $\angle DAF= 50^{\circ}$ . A few other angles can also be easily derived from angles of a triangle sum to $180^{\circ}$ , but not the one we want.

Apply the law of sines on $\triangle BAD$ and $\triangle CAF$ which have an equal side and can be combined. Calling $\frac{BD}{FC}=\frac{DE}{EF}=a$ allows us to solve for $a$ .

$a=\frac{\sin{30^{\circ}}\sin{120^{\circ}}}{\sin{20^{\circ}}\sin{110^{\circ}}}$

Call $\angle DAE=\alpha$ so $\angle AEF=(50^{\circ}-\alpha)$ then apply the law of sines on these two triangles which have a side in common. Combine as before but this time rewrite the result as

$\frac{\sin{\alpha}}{\sin{(50^{\circ}-\alpha)}}=\frac{a \sin{70^{\circ}}}{\sin{60^{\circ}}}$

Finally, substitute $a$ into this and reduce, since $\sin{70^{\circ}}=\sin{110^{\circ}}$ and $\sin{120^{\circ}}=\sin{60^{\circ}}$ and you're left with

$\frac{\sin{\alpha}}{\sin{(50^{\circ}-\alpha)}}=\frac{\sin{30^{\circ}}}{\sin{20^{\circ}}}$

So, by inspection $\alpha=\angle DAE = \boxed{30}^{\circ}$

Given < BAC = 100, AB = AC implies that <ACF = < ABD = 40. Since <CAF = 20, < AFC = 120, < AFE = 60. Let x = <BAD. Then <DAF = x + 20.In triangle DAF, < ADF = 100 - x, and < BDA = 80 + x. Then in triangle ABD, 180 = x + 40 + 80 + x, or 180 = 2x + 120, and x = 30.Then <DAF = 100 - 30 - 20 = 50 = x + 20, so <DAE = 30. Ed Gray