The point $P$ on the given segment $AB$ is obtained in the following steps: 1) From $B$, measure the segment $BC=AB/2$ perpendicularly to $AB$; 2) with the line perpendicular to $AC$ through $C$, intersect the line $AB$ to get the point $D$; 3) from $D$, measure the segment $DE$ equal in length and direction to $AB$; 4) with the line perpendicular to $AB$ through $E$, intersect the line $AC$ to get the point $F$; 5) with the line parallel to $DF$ through $C$, intersect $AB$ to get $P$. - In what ratio does $P$ divide the segment $AB$?
The triangles $ABC$ and $CBD$ are similar because their corresponding sides are perpendicular to each other. Therefore, $BD:BC = BC:BA = 1:2$, so $BD = AB/4$, $AD = 5AB/4$, and $AE = 9AB/4$. From the similar triangles $APC$ and $ADE$, and the similar triangles $ABC$ and $AEF$,

$$
AP:AD = AC:AF = AB:AE = 4:9
$$

thus $AP = 4AD/9 = 5AB/9$, hence $AP:PB = 5:4$.

![](https://cdn.mathpix.com/cropped/2024_05_02_3a3de0f46c7b3ae3d999g-1.jpg?height=390&width=618&top_left_y=461&top_left_x=748)

Péter Sziklai (Budapest, Bláthy O. Erősár. Ip. Techn. II. o. t.)

Remark. We can also use the fact that the triangles $PBC$ and $DEF$ are in similar positions: $EF = 9BC/4$, so $PB = 4DE/9 = 4AB/9$, $AP = 5AB/9$.