Triangle inscribed inside a circle: prove that $abc = 4 \times area \times R$ The solution to Putnam 2000 A5 uses this formula, for which the following proof is given:
(source: https://mks.mff.cuni.cz/kalva/putnam/psoln/psol005.html)



Let the sides (of triangle $ABC$) have lengths $a, b, c$ as usual. The question suggests that we use some relationship of the form $abc = constant \times R$. ...

To prove the relation, let $O$ be the centre of the circumcircle. Project $AO$ to meet the circle again at $K$. Let $AH$ be the altitude. Then angle $ABC = \angle AKC$, so triangles $ABH$ and $AKC$ are similar. Hence $\frac{AB}{AH} = \frac{AK}{AC}$ or $\frac{c}{AH} = \frac{2R}b$. Hence $abc = 2R·a·AH = 4\Delta R$.



The bold part confuses me. How does $ABC = AKC$? $K$ is dependent wholly on $O$ and $A$ whereas $B$ is independent of both. And if $ABC = AKC$, how does that lead to $ABH \sim AKC$?
Angles intercepting the same arc are equal. Both angles are intercepting $\widehat{AC}$.
Area of the inscribed $ \triangle = \frac 12 a.b\sin C$

Try to draw the plane and we will get $\sin C = \frac{c}{2R}$