Reflect a triangle's one median over the internal angle bisector from the same vertex.

a) In what ratio does each reflection divide the opposite side?

b) Is it true that the three reflections lie on a straight line?
a) Let the midpoint of side $a$ be $A_{1}$, the midpoint of side $b$ be $B_{1}$, and the reflection of the median $A A_{1}$ over the angle bisector $A D$ intersect side $a$ at point $A_{0}$. The other notations can be found in the diagram.

![](https://cdn.mathpix.com/cropped/2024_05_02_cc16a2f1550d30e098c6g-1.jpg?height=355&width=585&top_left_y=272&top_left_x=759)

We want to determine the ratio $\frac{B A_{0}}{A_{0} C}$. The line through $A_{0}$ parallel to $A C$ intersects side $c$ at point $E$. Since segments $A_{1} B_{1}$ and $A B$ are parallel, the angles marked with two arcs are equal. By symmetry with respect to the angle bisector $A D$, the angles marked with one arc are also equal. Therefore, triangles $A E A_{0}$ and $A B_{1} A_{1}$ are similar,

$$
\frac{A_{0} E}{E A}=\frac{A_{1} B_{1}}{B_{1} A}
$$

This can also be written as:

$$
\frac{A_{0} E}{c-E B}=\frac{A_{1} B_{1}}{B_{1} A}=\frac{\overline{2}}{\frac{b}{2}}=\frac{c}{b}
$$

It is clear that triangles $E B A_{0}$ and the original triangle $A B C$ are similar, and therefore

$$
\frac{A_{0} E}{B A_{0}}=\frac{b}{a} \quad \text{and} \quad \frac{E B}{B A_{0}}=\frac{c}{a}
$$

From these two equalities, we can express $A_{0} E$ and $E B$, and substitute them into (1):

$$
\frac{B A_{0} \cdot \frac{b}{a}}{c-B A_{0} \cdot \frac{c}{a}}=\frac{c}{b}
$$

From this,

$$
\frac{B A_{0}}{a-B A_{0}}=\frac{c^{2}}{b^{2}}
$$

or

$$
\frac{B A_{0}}{A_{0} C}=\frac{c^{2}}{b^{2}}
$$

Thus, the answer to the first part of the problem is that a reflection divides the opposite side in the ratio of the squares of the sides emanating from the vertex.

b) The answer to the second part of the problem is affirmative. It is clear that $A_{0}$ and the similarly defined $B_{0}$ and $C_{0}$ are interior points of the sides of the triangle, and based on the result from part a):

$$
\frac{B A_{0}}{A_{0} C} \cdot \frac{C B_{0}}{B_{0} A} \cdot \frac{A C_{0}}{C_{0} B}=\frac{c^{2}}{b^{2}} \cdot \frac{a^{2}}{c^{2}} \cdot \frac{b^{2}}{a^{2}}=1
$$

From this, by the converse of Ceva's theorem, it follows that the three lines intersect at a single point.