Area of triangle ABC inside circle

Question

Consider the following diagram:

$AB+AD=DE$, $\angle BAD= 60$, and $AE$ is $6$. How do we find the area of the triangle $ABC$?

Answer 1

From the above picture,

$x+y=z$ , $ y+z=6$ and $\frac x {y+z} = \cos 60^\circ = \frac 1 2$

After calculation, $x=3$, $y=\frac 3 2$ and $z = \frac 9 2$

$\angle ABE = 90^\circ$ and $\angle BAD = 60^\circ$

So, $\angle AEB = 30^\circ = \angle ACB$ (properties of a circle)

Now, $\cos 60^\circ = \frac {x^2 + y^2 - w^2}{2xy} = \frac 1 2$

After calculation, $w=DB=\frac {3 \sqrt 3}{2}$

$$\cos \angle ADB=\frac{y^2+w^2-x^2}{2yw}=\frac{(\frac 3 2)^2+(\frac {3 \sqrt 3}{2})^2-3^2}{2 \frac 3 2\frac {3 \sqrt 3}{2}}=0$$

So, $\angle ADB = 90^\circ$ and $\angle ABD = 30^\circ = \angle ACB$

So, $\triangle ABC $ is an isosceles triangle.

And, $CD = BD = \frac {3 \sqrt 3}{2}$

So, $\triangle ABC = \frac 1 2 \cdot \frac 3 2 \cdot (\frac {3 \sqrt 3}{2} + \frac {3 \sqrt 3}{2}) = \frac {9 \sqrt 3}{4}$