Library / Saudi Arabian Mathematical Competitions /350 of 395

Saudi Arabian Mathematical Competitions, 2015

Geometry Difficulty 6.8 National Olympiad Prove it Saudi Arabia

Let $ABC$ be a triangle and $D$ a point on the side $BC$ . The tangent line to the circumcircle of the triangle $ABD$ at the point $D$ intersects the side $AC$ at $E$ . The tangent line to the circumcircle of the triangle $ACD$ at the point $D$ intersects the side $AB$ at $F$ . Prove that the point $A$ and the circumcenters of the triangles $ABC$ and $DEF$ are collinear.

Solution

Let $\alpha = \angle BAC$ , $\beta = \angle CBA$ , $\gamma = \angle ACB$ and $\theta = BAD$ . Because line $DE$ is tangent to the circumcircle of triangle $ABD$ , we have $\angle EDA = \angle DBA = \beta$ . Because line $DF$ is tangent to the circumcircle of triangle $ADC$ , we have $\angle ADF = \angle ACD = \gamma$ .

Therefore
$\angle EDF + \angle FAE = \angle EDA + \angle ADF + \angle BAC = \gamma + \beta + \alpha = 180^\circ.$
This proves that quadrilateral $AFDE$ is cyclic and hence $\angle EFA = \angle EDA = \beta$ and $\angle AEF = \angle ADF = \gamma$ . This proves that sides $EF$ and $BC$ are parallel. Let $O$ and $O'$ be circumcenters of triangle $ABC$ and $AFE$ , respectively. We have
$\angle FAO' = 90^\circ - \angle AEF = 90^\circ - \gamma = \angle BAO.$
This proves that $A$ , $O$ and $O'$ are collinear.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.