In the isosceles triangle $ABC$, the midpoint of the base $BC$ is $F$, and a point $D$ lies on the segment $BF$. The perpendicular from $D$ to $BC$ intersects $AB$ at $M$, and the line through $D$ parallel to $AB$ intersects $AC$ at $P$. Determine the ratio of the areas of triangles $AMP$ and $ABC$ as a function of $k = BD : BC$.

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Solution. The angles at vertex $A$ of triangles $A M P$ and $A B C$ are common, so if this angle is denoted by $\alpha$, then using the equality $A B = A C$, the ratio of their areas is

$$
\frac{T_{A M P}}{T_{A B C}}=\frac{\frac{A M \cdot A P \sin \alpha}{2}}{\frac{A B \cdot A C \sin \alpha}{2}}=\frac{A M \cdot A P}{A B \cdot A C}=\frac{A M}{A B} \cdot \frac{A P}{A B}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_a0efd6770ae437d63e43g-1.jpg?height=518&width=458&top_left_y=413&top_left_x=823)

Since segments $A B$ and $P D$ are parallel, by the theorem of parallel intercepts, $A P: A C = B D: B C = k$, and using the equality $A B = A C$ again, we get $A P = k \cdot A B$. Triangle $A B C$ is isosceles, so $A F$ is parallel to $M D$, and using the theorem of parallel intercepts again, we find that

$$
A M = A B - B M = A B - A B \cdot \frac{B D}{B F} = A B - A B \cdot \frac{2 B D}{B C} = A B(1 - 2 k)
$$

Therefore, the ratio of the areas of the two triangles is

$$
\frac{A M}{A B} \cdot \frac{A P}{A B} = (1 - 2 k) k = k - 2 k^{2}
$$