Find the Area a Kite Inscribed in a Circle

Question

The question is as follows:

$A$, $B$, $C$ and $D$ lie in alphabetical order on a circle so that ABCD forms a kite. $AB = DA = 8 cm$ and $BC = CD = 13 cm$. Find the area of the kite $ABCD$.

I thought that I might be able to use $AC$ as the diameter of the circle, therefore, $\angle ADC$ would be a right angle. Using that I thought that I can use the formula for finding the area of a triangle using sine ($\frac{1}{2}ab \times \sin C$) to find the area of the two congruent triangles. However, the answer that I get does not match with the correct answer. What am I doing incorrectly? Any help will be greatly appreciated!

Answer 1

An overkill solution:

We know that the area of kite is $ef/2$ where $e,f$ are diagonals. By Ptolomey theorem we have:

$$ 2A =ef = ac+bd = 2ac = 2\cdot 104$$ so $A= 104$.

Answer 2

I leave it as an exercise to find why ($M$ is the point where the perpendicular diagonals of the kite cut)

$$ \frac{1}{BM^2}= \frac{1}{8^2} + \frac{1}{13^2} $$

is correct ( compare proportionate sides of similar triangles )

Now as you have done for total kite area

$$ (AC \cdot BD/2 \cdot \frac12) \cdot 2 $$

$AC$ is hypotenuse etc.