9,10,11 |

Authors: Bverlov S.L., $\underline{\text { Yakubov A. }}$.

Given a parallelogram $A B C D$, where $A B<A C<B C$. Points $E$ and $F$ are chosen on the circumcircle $\omega$ of triangle $A B C$ such that the tangents to $\omega$ at these points pass through point $D$; moreover, segments $A D$ and $C E$ intersect. It turns out that $\angle A B F=\angle D C E$. Find the angle $A B C$.

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Since $D$ lies outside $\omega$, angle $A B C$ is acute. Let $A^{\prime}$ be the second intersection point of $D C$ and $\omega$. Since $B C > A C$, we have

$\angle D C A = \angle C A B > \angle C B A = \angle D A ' A$; hence, $A^{\prime}$ lies on the extension of segment $D C$ beyond point $C$. Notice that $-E C A^{\prime} = 2(180^{\circ} - \angle E C A^{\prime}) = 2 \angle E C D = 2 \angle A B F = -A C F$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8b1cc902b566a3f27ee1g-44.jpg?height=546&width=652&top_left_y=2065&top_left_x=703)

Let $l$ be the angle bisector of $\angle E D F$. Since $D E$ and $D F$ are tangents, line $l$ passes through the center $O$ of circle $\omega$. Reflecting over $l$, $\omega$ maps to itself. Since $-E C A^{\prime} = -A C F$, points $A$ and $A^{\prime}$ map to each other under this reflection. Therefore, $\angle D A A^{\prime} = \angle D A^{\prime} A$. On the other hand, since point $A^{\prime}$ lies on $\omega$, $\angle A A^{\prime} C = \angle A B C = \angle A D A^{\prime}$. Thus, all three angles of triangle $D A A^{\prime}$ are equal, implying $\angle A B C = \angle A D A^{\prime} = 60^{\circ}$.

## [ Orthocenter and Orthic Triangle [ Inscribed Angle Equals Half of Central Angle ] Problem 65361 Topics: [Angle Between Two Chords and Two Secants] [ Angles Subtending Equal Arcs and Equal Chords ] [ Sum of Angles in a Triangle. Exterior Angle Theorem. ] Difficulty: $4-$ Grades: $8,9,10,11$

Author: Binkov A.d:

A circle passing through vertices $A, B$ and the orthocenter of triangle $A B C$ intersects sides $A C$ and $B C$ at internal points.

Prove that $\angle C = 60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8b1cc902b566a3f27ee1g-45.jpg?height=446&width=600&top_left_y=1089&top_left_x=729)

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