## Problem 2.

Consider triangle $A B C$ with $\angle A=30^{\circ}$ and $\angle B=80^{\circ}$. On sides $A C$ and $B C$, points $D$ and $E$ are taken such that $\angle A B D \equiv \angle D B C$ and $D E \| A B$.

Determine the measure of angle $\angle E A C$.

![](https://cdn.mathpix.com/cropped/2024_06_07_cc89f2f54d84ee4a810cg-3.jpg?height=376&width=662&top_left_y=1010&top_left_x=688)
Solution. Let $A M \perp B C$ and $A F$ be the bisector of angle $\angle M A C$, points $M$ and $F$ being located on side $B C$. Clearly, $\angle B A M = \angle M A F = 10^{\circ}$, so $A M$ is both an altitude and a bisector in triangle $A B F$. We deduce that $B M = M F$. .................. 1p

Applying the Angle Bisector Theorem in triangle $A B C$ we get that $\frac{C D}{D A} = \frac{B C}{A B} \ldots \ldots \ldots \mathbf{p}$

Applying the Angle Bisector Theorem in triangle $A M C$ we get that $\frac{C F}{M F} = \frac{A C}{A M}$. It follows

![](https://cdn.mathpix.com/cropped/2024_06_07_cc89f2f54d84ee4a810cg-3.jpg?height=109&width=1577&top_left_y=1840&top_left_x=228)

But $\frac{1}{2} \frac{A C}{A M} = \frac{1}{2} \frac{A C \cdot A B}{A M \cdot A B} = \frac{A C \cdot A B \cdot \sin 30^{\circ}}{A M \cdot A B} = \frac{2 S_{A B C}}{A M \cdot A B} = \frac{A M \cdot B C}{A M \cdot A B} = \frac{B C}{A B} . \quad \mathbf{3 p}$

Thus, $\frac{C D}{D A} = \frac{C F}{F B}$, hence lines $D F$ and $A B$ are parallel.

This means that points $E$ and $F$ coincide, so $\angle E A C = 10^{\circ}$. 1p

![](https://cdn.mathpix.com/cropped/2024_06_07_cc89f2f54d84ee4a810cg-4.jpg?height=271&width=268&top_left_y=130&top_left_x=858)

## First Selection Barrier for OBMJ Iași, April 19, 2022  Solutions and Marking Schemes