## A Line in Triangle Through the Circumcenter

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

Let O be the circumcenter of ΔABC. A line through O intersects the sides AB and AC at M and N, respectively. Let S and R be the midpoints of BN and CM. respectively. Prove that ∠ROS = ∠BAC.

4 December 2015, Created with GeoGebra

Solution ### 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

∠BOR + ∠COS = ∠BAC.

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

∠BOC = 2∠BAC.

Finally,

 ∠ROS = ∠BOC - (∠BOR + ∠COS) = 2∠BAC - ∠BAC = ∠BAC.

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