# Triangle inscribed inside a circle: prove that abc = 4 * area * 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 x 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 AB/AH = AK/AC or c/AH = 2R/b. Hence abc = 2R·a·AH = 4ΔR.

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 ~ AKC?

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Inscribed Angle Theorem? –  Blue Jul 30 at 23:08
The bold part is in fact the result of a theorem called "angles in the same segment". Or equivalently, "equal arcs, then equal angles". –  Mick Jul 31 at 5:28
Note also that angle ABH = 90 degrees (AH is the altitude from A to BC) and angle ACK = 90 degrees (angles in semi-circle). Therefore, the two triangles are similar (AAA). –  Mick Jul 31 at 8:25
@Mick Thanks for the comment. But I am still confused. Don't you mean AHB not ABH? AHB = ACK = 90 and ABC = AKC. But how does that make ABH ~ AKC? In order to prove similarity we need ABH = AKC but we have ABC = AKC instead. imgur.com/2oUBqPr –  user1299784 Jul 31 at 21:28
@user1299784 The confusion comes from you and the author (and also me) are drawing different pictures. It will be quite clear if you re-draw your picture and let your angle B be acute (instead of obtuse). Then, H will lie inside the circle (instead of outside). The next question is then “why an obtuse triangle won’t work?” The answer is “it will also work" but the proof will run differently. –  Mick Aug 1 at 2:24

Angles intercepting the same arc are equal. Both angles are intercepting $\widehat{AC}$.