A Line in Triangle Through the Circumcenter

The applet below illustrates the following statement I came across at the WOOT of the site:

Solution

The problem is easily solved with a variant of Pascal's Hexagram Theorem. This is how.

Let BB_t and CC_t be two diameters of the circumcircle. First, observe that a homothety with center B and coefficient 2 maps O to B_t and R to N. It follows that OR||NB_t. Similarly, OS||MC_t. Let B_t and MC_t meet at K. The proof is based on the fact that K lies on the circle. This may not be immediately obvious and I deal with this assertion on a separate page.

The inscribed angles at B_t and C_t add up to the inscribed angle at A, i.e. ∠BAC, implying

On the other hand, since ∠BOC is the central angle subtended by the same arc as the inscribed ∠BAC, we have

Finally,

The fact that K lies on the circle is a special reformulation of Pascal's theorem.