## Task A-2.7.

Let $A B C D$ be a parallelogram and let $S$ be the intersection of its diagonals. The angle bisector of $\angle A D C$ bisects the segment $\overline{A S}$ and intersects the line $B C$ at point $E$. Determine the ratios $|B E|:|B C|$ and $|A B|:|B C|$.
## Solution.

Let line $D E$ intersect diagonal $\overline{A C}$ at point $F$.

![](https://cdn.mathpix.com/cropped/2024_05_30_abcf4bcd4243d7db1a46g-12.jpg?height=643&width=793&top_left_y=1640&top_left_x=563)

The diagonals of a parallelogram bisect each other, so point $S$ is the midpoint of $\overline{A C}$. Since point $F$ is the midpoint of $\overline{A S}$, it follows that $|A F|:|F C|=1: 3$.

Angles $\varangle A D E$ and $\varangle D E C$ are congruent (angles formed by the intersection of line $D E$ with parallel lines $A D$ and $C E$), as are angles $\varangle A F D$ and $\varangle C F E$ (vertical angles).

This means that $|A D|:|C E|=|A F|:|F C|$, and thus $|C E|=3|A D|$.

1 point

Since $A B C D$ is a parallelogram, $|A D|=|B C|$, so $|B E|=|E C|-|B C|=$ $3|B C|-|B C|=2|B C|$, and the desired ratio $|B E|:|B C|=2: 1$.

Furthermore, from $\varangle A D E=\varangle D E C$, using the fact that $D E$ is the angle bisector of $\varangle A D C$, i.e., $\varangle A D E=\varangle E D C$, we conclude that $\varangle E D C=\varangle D E C$.

1 point

Thus, triangle $D C E$ is isosceles, and $|C D|=|C E|$.

1 point

Since $|A B|=|C D|$, and $|C E|=3|A D|=3|B C|$, we see that $|A B|=3|B C|$,

2 points i.e., $|A B|:|B C|=3: 1$.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

3rd grade - high school - A variant
January 4, 2013.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.