What is the area of the triangle here? 
We are given that the angle of $BAD$ is $2\alpha$ and the angle of $DAC$ is $\alpha$. $|AC| = 10$, $|BD| = 6$, $|DC| = 5$ units. Find the area of the triangle $ABC$.
The answer would be $33$. We need to show that if we drop an altitude from $A$ to $BC$ at point $E$, $|AE| = 6, |EC| = 8$. Somehow $\triangle AEC$ becomes $6$-$8$-$10$ triangle. Since $|AE| = 6$ and $|BC|=11$, the area becomes $33$.
But how can we prove that?

Draw $AE$, the angle bisector of $\angle BAD$. Now $\angle A$ is divided into $3$ equal angles. Say, $DE = x$. As $AD$ bisects $\angle CAE$, using angle bisector theorem,
$ \displaystyle AE = 2 DE = 2x$
Using formula for the length of angle bisector,
$AD^2 = AE \cdot AC - DE \cdot CD = 15x$
Now using the fact that $AE$ is angle bisector of $\angle BAD$,
$ \displaystyle \frac{AB}{AD} = \frac{BE}{DE} \implies \frac{AB}{\sqrt{15x}} = \frac{6-x}{x}$
Or, $~AB \cdot AD = 15 (6-x)$
Again using formula for the length of angle bisector, $AE^2 = AB \cdot AD - AD \cdot BE$
Or, $4 x^2 = 90 - 15x - x (6-x)$
Or, $x^2 + 7x - 30 = 0$
Or, $(x-3) (x + 10) = 0$
The only valid solution is $x = 3$. That shows $AE = 2x = 6, CE = 5 + x = 8$ and we conclude $AE \perp BC$
$\therefore [ABC] = 33$
