In triangle $ABC$, $AB=6$ and $BC=10$ units. The foot of the angle bisector from $B$ is connected to the midpoint of side $AB$. How far from $B$ does this line intersect the line $BC$?
I. solution. Let the midpoint of $AB$ be denoted by $F$, the foot of the angle bisector from $B$ by $T$, the intersection of $FT$ and $BC$ by $M$, and the reflections of $C$ and $T$ over $F$ by $C'$ and $T'$, respectively.

By the Angle Bisector Theorem, $AT: TC = AB: BC = 6: 10$, so we can assume that $AT = 3x$ and $TC = 5x$. The reflection of $AT$ over $F$ is $BT'$, hence $BT' = 3x$ and $BT' \| AT$. We can apply the theorem of parallel intercepts:

![](https://cdn.mathpix.com/cropped/2024_05_02_833505f509b0bb3a724fg-1.jpg?height=810&width=2025&top_left_y=335&top_left_x=39)

(The intersection point $M$ is always on the extension of segment $CB$ beyond $B$ due to $\frac{AT}{TC} < \frac{AF}{FB}$!|17 From (1), a simple calculation shows that $MB = 15$ units.

Berki Csaba (Székesfehérvár, Teleki B. Gimn., I. o.t.)

II. solution. Using the notations from the previous solution. According to Menelaus' Theorem (see, for example, Geometric Problems Collection I., problem 1260)

$$
\frac{AF}{FB} \cdot \frac{BM}{MC} \cdot \frac{CT}{TA} = -1
$$

From this, using the Angle Bisector Theorem, we immediately obtain the (1) equality.

Becker Johanna (Budapest, Árpád Gimn., II. o.t.)[^0]


[^0]:    ${ }^{1}$ Most of our contestants did not see this.