Problem 1447

1. Setup and Definitions:
- Let $AA_0$ be the altitude of the isosceles triangle $ABC$ with $AB = AC$.
- Let $\gamma$ be a circle centered at the midpoint $M$ of $AA_0$ that touches $AB$ and $AC$.
- Let $X$ be an arbitrary point on the line $BC$.
- We need to prove that the tangents from $X$ to $\gamma$ cut congruent segments on lines $AB$ and $AC$.

2. **Properties of the Circle $\gamma$**:
- Since $\gamma$ is centered at the midpoint $M$ of $AA_0$ and touches $AB$ and $AC$, it is the incircle of the right triangle $AMA_0$.
- The radius of $\gamma$ is the distance from $M$ to $AB$ or $AC$.

3. **Tangents from $X$ to $\gamma$**:
- Let the tangents from $X$ to $\gamma$ touch $\gamma$ at points $P$ and $Q$.
- By the properties of tangents from a point to a circle, $XP = XQ$.

4. Reflection and Symmetry:
- Reflect $P$ and $Q$ over the line $AA_0$ to get points $P'$ and $Q'$ respectively.
- Since $\gamma$ is symmetric with respect to $AA_0$, the tangents from $X$ to $\gamma$ will also be symmetric with respect to $AA_0$.

5. **Congruent Segments on $AB$ and $AC$**:
- Let $P$ and $Q$ intersect $AB$ and $AC$ at points $S$ and $R$ respectively.
- By the symmetry of the setup, the segments $AS$ and $AR$ are congruent to $AP'$ and $AQ'$ respectively.

6. Using Menelaus' Theorem:
- Consider the triangle $ABC$ with transversal $PQR$.
- By Menelaus' theorem, we have:
$$\frac{AS}{SB} \cdot \frac{BR}{RC} \cdot \frac{CQ}{QA} = 1$$
- Since $AS = AR$ and $BS = CR$, the segments $AS$ and $AR$ are congruent.

7. Conclusion:
- The tangents from $X$ to $\gamma$ cut congruent segments on lines $AB$ and $AC$.

$\blacksquare$