Suppose we have an inscribed isosceles triangle in a circle.

Question

$ABC$ is an isosceles triangle inscribed in a circle with centre $O$.Suppose the top vertex is $A$, the right vertex is $B$ and the left vertex is $C$. Also suppose $AC=AB$. Furthermore, the diameter $AD$ goes through $O$ and intersect $CB$ at $E$. IF $AC= \sqrt{\frac{15}{2}}$ and $OE =1$, estimate the radius of the circle.

I drew this out but cannot come up with a way to get the radius much less estimate it.

I do see similar triangles but there’s still two unknowns in the equation.

Any nudging would be appreciated.

Answer 1

Hint:  use Pythagoras' theorem twice, then eliminate $\,CE\,$ between the following to find $\,r\,$:

$$ OE^2+CE^2 = r^2 \\ (OE+r)^2 + CE^2 = AC^2 $$

Answer 2

Let $r$ be the length of radius of the circle. When we connect $O$ and $B$, we have $|OB| = r$. So, by Pythagoras Theorem $|EB| = |EC| = \sqrt{r^2-1}$ ($|EB| = |EC|$ because we have isosceles tirangle with |AB| = |AC|).

Now, notice the right angle triangle $\Delta AEC$. We have $|AE| = r+1$, $|CE| = \sqrt{r^2-1}$ and $|AC| = \sqrt{\frac{15}{2}}$. So, by Pythagoras Theorem, we have $$|AE|^2+|CE|^2 = |AC|^2$$ $$(r+1)^2+(\sqrt{r^2-1})^2 = \frac{15}{2} \implies 4r^2+4r-15 = 0 \implies (2r-3)(2r+5) = 0$$ $$\implies r = \frac{3}{2}$$