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?
 A: 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 $AE \perp BC$
$\therefore [ABC] = 33$
A: 
Hints:
Draw perpendicular bisector of BC, it touches the circumcircle at G. AG is bisector of angle BAC. If AH is altitude from A and O locates on AD, then we have:
$$\widehat {HAG}=\widehat{AGO}=\widehat{GAD}$$
which means that:
$$\angle ABH=\angle HAD=\angle DAC $$
Only in such orientation $\widehat {BAD}=2 \widehat{DAC}$
H is midpoint of BD and $\triangle ABD$ is isosceles. Also we have:
$$\cos \widehat {ACD}=\frac{5+\frac 62}{10}=0.8$$
In right angled triangle OFC we have:
$$\cos \widehat {FCD}=0.8=\frac {FC}{DC=5}\Rightarrow FC=4\rightarrow AF=10-4=6$$
Right angled triangles AHD and ADF are equal therefore:
$$AH=AF=6$$
$$S_{ABC}=\frac {6\times 11}2=33$$
A: In the following figure, we have

(1) Circle C with radius CA is drawn.
(2) A and A’ are two points on the circumference with A being higher than A’ w.r.t. the reference line CDB.
By considering the line DC, we have $\angle DAC \ge \angle DA’C$ and also w.r.t. BD, we have $\angle BA’D \ge \angle BAD$.
It should be clear that there is a point Y on BD and midway between B and D such that [ABC ] = 33. In addition, applying the converse of the angle bisector theorem, we can show $\beta : \alpha = 2 : 1$.

The question is:- “is this the only configuration?” This is because on the arc AX, some other points (like) A’ could be a possible candidate.
The facts shown above show that A’ is not possible because at A’, $\alpha’$ is getting smaller than $\alpha$ while $\beta’$ is getting larger than $\beta$ such that they can never meet the 2 :1 requirement.
A: Let area of $\triangle BAD = A_1$ and area of $\triangle CAD = A_2$. Let $AD = m$ and $AB = x$.
As they share a common height, the ratio $\frac{A_1}{A_2} = \frac 65$
Also, using an alternative formula for triangular area, $\frac{A_1}{A_2} = \frac{\frac 12xm\sin 2\alpha}{\frac 12(m)(10)\sin \alpha} = \frac{x\cos\alpha}{5}$ (where a double angle identity was used).
Equating the two expressions and rearranging, $\cos \alpha = \frac 6x$.
Now apply cosine rule on $\triangle ABC$, to give $11^2 = x^2 + 10^2 - 2(x)(10)\cos 3\alpha$.
Using a triple angle identity ($\cos 3\alpha = 4\cos^3 \alpha - 3\cos\alpha$),
$x^2 - 20x(4(\frac 6x)^3 - 3(\frac 6x)) - 21 = 0$
After a bit of algebra and setting $y = x^2$ to reduce the quartic to a quadratic, we get:
$y^2 + 339y - 17280 = 0$.
Solving for the positive root, $y = 45 \implies x  =\sqrt{45}$.
At this point, we can proceed to find the area of $\triangle ABC$ in a couple of ways. One way is to apply Heron's formula using the three known sides. I instead opted to apply the previously used formula, so that the area $A = \frac 12(x)(10)\sin 3\alpha$.
From $\cos \alpha = \frac 6x$ we get $\sin \alpha = \frac 3{\sqrt{45}}$. Applying a triple angle formula ($\sin 3\alpha = 3\sin\alpha - 4\sin^3\alpha$),
$A = \frac 12(x)(10)(\frac 9{\sqrt{45}} - 4(\frac{27}{45\sqrt{45}})) = 5\sqrt{45}(\frac 9{\sqrt{45}} - \frac{12}{5\sqrt{45}}) = 45- 12 = 33$.
So the answer is $33$.
