Radius of the circumscribed circle of an isosceles triangle

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.

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:

$$R=\frac{p}{4\cos^2\left(\frac{\alpha}{2}\right)\sin\left(\frac{\alpha}{4}\right)}$$

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.