We folded a paper in the shape of an isosceles triangle such that the vertex $A$ of the base landed at the midpoint of the side $BC$. In this case, one end of the fold line divides the side $AC$ in the ratio $p: q$. In what ratio does the other end of the fold line divide the base $AB$?

Following Magyar Hírlap
During the folding, vertex $A$ lands on point $F$, the midpoint of side $BC$, so the fold line coincides with segment $AF$, which is the median to side $BC$. Let's denote the intersection points of the perpendicular bisector of $AF$ with segments $AC$, $AF$, and $AB$ as $K$, $L$, and $M$, respectively.

First, let's examine the ratio $p: q$ under which $M$ is an internal point of base $AB$. If the line $KL$ passes through $B$ (1st figure), then, denoting the intersection of the line through $F$ parallel to $KL$ and $AC$ as $G$, the intercept theorem gives us $AK: KG = AL: LF = 1: 1$ and $KG: GC = BF: FC = 1: 1$. Therefore, $p: q = AK: KC = AK: (KG + GC) = 1: 2$. If the ratio $p: q$ decreases, or if the vertex $C$ moves away from the base while $AB$ is fixed, then the line $KL$ intersects the line $AB$ outside the segment $AB$ - on its extension beyond $B$.

![](https://cdn.mathpix.com/cropped/2024_05_02_bc001e32fc2466617504g-1.jpg?height=561&width=284&top_left_y=478&top_left_x=324)

1st figure

![](https://cdn.mathpix.com/cropped/2024_05_02_bc001e32fc2466617504g-1.jpg?height=558&width=379&top_left_y=1171&top_left_x=602)

2nd figure

![](https://cdn.mathpix.com/cropped/2024_05_02_bc001e32fc2466617504g-1.jpg?height=635&width=377&top_left_y=1081&top_left_x=1465)

3rd figure

We can assume that the equalities (1) and (2) are also true when $q \geq 2p$, and thus the final result is the same. The proof of this can be read from the 3rd figure.