## 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

4 December 2015, Created with GeoGebra

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Copyright © 1996-2017 Alexander Bogomolny### 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

∠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

- Munching on Inscribed Angles
- More On Inscribed Angles
- Inscribed Angles
- Tangent and Secant
- Angles Inscribed in an Absent Circle
- A Line in Triangle Through the Circumcenter
- Angle Bisector in Parallelogram
- Phantom Circle and Recaptured Symmetry
- Cherchez le quadrilatere cyclique
- Cyclic Quadrilateral, Concurrent Circles and Collinear Points
- Parallel Lines in a Cyclic Quadrilateral
- Reim's Similar Coins I
- Reim's Similar Coins II
- Reim's Similar Coins III
- Reim's Similar Coins IV
- Pure Angle Chasing
- Pure Angle Chasing II
- Pure Angle Chasing III

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Copyright © 1996-2017 Alexander Bogomolny62618450 |