Radius of the circumscribed circle of an isosceles triangle An isosceles triangle $ABC$ is given $(AC=BC).$ The perimeter of $\triangle ABC$ is $2p$, and the base angle is $\alpha.$ Find the radius of the circumscribed circle $R$.

$$R=\frac{p}{2\sin\alpha(1+\cos\alpha)}$$


Let $CD=2R.$ The triangle $BCD$ is a right triangle and we have $\angle BAC=\angle ABC=\angle BDC=\alpha.$
I am not sure how to approach the problem. It's really hard for me to solve problems like this. Can you give me a hint and some thoughts on the problem?

Another simple approach. Let $x=AC=BC$.  Then
$$2p=AC+BC+2AH\\=2x+2x\cos\alpha$$
and
$$R=\frac 12 CD=\frac 12 \frac{BC}{ \sin \alpha} = \frac{x}{2 \sin \alpha}$$
Now you can complete the solution by a simple substitution.
