A circle is drawn on one side of a triangle. Prove the radius of that circle.

Question

Question: In $\triangle ABC$, $\angle C = θ$, $|CB| = a$ and $|CA| = b$. The semi-circle with the diameter lying on the side $[AB]$ touches the side $[CA]$ and the side $[CB]$. Prove the radius of the semi-circle is equal to

$$\dfrac{ab\cdot \sin θ}{a+b}$$

I found that $c = \sqrt{a^2+b^2-2ab\cos θ}$, but I am not sure where to go on from there. The cosine formula gives you c using cosine, but the number given in the question uses sine.

Answer 1

Area of $ABC$ is

1. $\frac{1}{2}ab\sin\theta$.
2. area of $BCD$ + area of $ACD$ = $\frac12 aR+\frac12bR$.