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?
 A: 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.
A: Hint: Use following formula:
$$R=\frac{p}{4\cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right)\cos\left(\frac{\gamma}{2}\right)}$$
Where $\alpha$, $\beta$ and $\gamma$ are angles on vertices A, B and C respectively. $\alpha=\beta$ , therefore we have:
$$R=\frac{p}{4\cos^2\left(\frac{\alpha}{2}\right)\cos\left(\frac{\gamma}{2}\right)}$$
And also:
$$2\alpha+\gamma=\pi$$
$$\implies\frac{\gamma}{2}=\frac{\pi}{2}-\frac{\alpha}{4}$$
Finally:
$$R=\frac{p}{4\cos^2\left(\frac{\alpha}{2}\right)\sin\left(\frac{\alpha}{4}\right)}$$
A: The diameter is $$CD = 2R = \sqrt{BD^2 + BC^2}$$ by the Pythagorean theorem, since $\angle CBD$ is inscribed in a semicircle, thus is a right angle.
Now use the trigonometric properties to deduce that $$BH = BD \sin \alpha,$$ and $$ BH = BC \cos \alpha.$$  We also have $$BH + BC = p,$$ because this is half the perimeter of $\triangle ABC$.  Now all that is left is to eliminate $BH$, $BD$, and $BC$ from these four equations.
A: Using two known general expressions for the area of $\triangle ABC$
\begin{align} 
S&=\rho r
\tag{1}\label{1}
,\\
S&
=2\,R^2\sin\alpha\sin\beta\sin\gamma
=4\,R^2\sin^3\alpha\cos\alpha
\tag{2}\label{2}
,
\end{align}
and the expression for the inradius of $\triangle ABC$
in terms of its semiperimeter $\rho$,
\begin{align} 
r&=
\rho\tan\tfrac\alpha2\tan\tfrac\beta2\tan\tfrac\gamma2
=
\rho\tan^2\tfrac\alpha2\cot\alpha
\tag{3}\label{3}
,
\end{align}
we can find that
\begin{align} 
R&=
\frac1{2\sin\alpha}\,\sqrt{
\frac{\rho r}{\sin\alpha\cos\alpha}
}
\tag{4}\label{4}
,\\
R&=
\frac1{2\sin\alpha}\,\sqrt{
\frac{\rho^2 \tan^2\tfrac\alpha2\cot\alpha}{\sin\alpha\cos\alpha}
}
=\dots=
\frac\rho{2\sin\alpha+\sin2\alpha}
\tag{5}\label{5}
.
\end{align}
