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