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A Line in Triangle Through the CircumcenterThe 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 BBt and CCt be two diameters of the circumcircle. First, observe that a homothety with center B and coefficient 2 maps O to Bt and R to N. It follows that OR||NBt. Similarly, OS||MCt. Let Bt and MCt 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 Bt and Ct 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.
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