For reference: Calculate the area of triangle $ABC$; if $ED = 16$;
$AB = 10$ and $D = \angle15^o$(Answer:$20$)
My progress:
I didn't get much.
$\triangle ECD - (15^o, 75^o) \implies EC = 4(\sqrt6-\sqrt2), CD = 4(\sqrt 6+\sqrt 2)$
Incenter Th.
$\triangle ABD: \frac{AC}{CI}=\frac{10+BD}{16+EA}\\ S_{CDE} = 4(\sqrt6-\sqrt2)4(\sqrt6+\sqrt2) = 32$
I thought about closing the ABD triangle but as it's any other triangle, I didn't see much of an alternative
On
Reflect $B$ across $AC$, we get new point $B'$ on $AD$. Clearly triangles $ABC$ and $AB'C$ are congurent so we have to find an area of the later one. But this is just $AB'\cdot CF/2$. So we need to find $CF$. Let $G$ be the midpoint of $ED$, then $\angle EGC = 30^{\circ}$ and $GC = 8$. Since triangle $CFG$ is half of equlateral triangle we have $$CF = {CG \over 2}=4$$
The key is to recognize that since $\overline{AC}$ bisects $\angle DAB$, the altitudes from $C$ to $\overline{DE}$ and $\overline{AB}$ are congruent. Then since $AB$ is $5/8$ times $DE$, the area of $\triangle ABC$ must be $5/8$ times the area of $\triangle CDE$. You already have the the area of $\triangle CDE$, so the area of $\triangle ABC$ follows.