Blinkov Yu.A.

On the sides $AB$ and $CD$ of the square $ABCD$, points $K$ and $M$ are taken respectively, and on the diagonal $AC$ - point $L$ such that $ML = KL$. Let $P$ be the point of intersection of segments $MK$ and $BD$. Find the angle $KPL$.
The first method. Let $O$ be the intersection point of $AC$ and $BD$; $N$ be the midpoint of segment $KM$ (see Fig. a). Since the midpoint of a segment with endpoints on parallel lines lies on a line equidistant from them, $ON \parallel AK$ and $\angle AON = 45^{\circ}$. On the other hand, since $LN \perp MK$, the points $L, O, N$, and $P$ lie on the same circle, that is, $\angle LPN = 180^{\circ} - \angle LON = \angle AON = 45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-57.jpg?height=515&width=529&top_left_y=1998&top_left_x=90)

Fig. a

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-57.jpg?height=483&width=575&top_left_y=2011&top_left_x=753)

Fig. b

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-57.jpg?height=869&width=489&top_left_y=1824&top_left_x=1459)

Fig. c

The second method. Let $P'$ be the second intersection point of the circumcircles of triangles $AKL$ and $CML$ (see Fig. b). We will prove that $P'$ coincides with $P$. For this, it is sufficient to prove that: 1) points $K, P'$, and $M$ lie on the same line; 2) $P'$ lies on diagonal $BD$.

We will prove the second statement. From the equality of $LK$ and $LM$, it follows that $\angle LKP' = \angle LMP'$, then by the property of inscribed angles, $\angle LAP' = \angle LKP' = \angle LMP' = \angle LCP'$, that is, point $P'$ is equidistant from points $A$ and $C$, and therefore lies on diagonal $BD$. Thus, $\angle KPL = \angle KAL = 45^{\circ}$.

The third method. Let $S$ be the intersection point of the perpendicular bisector of segment $KM$ and line $AB$, and $N$ be the intersection point of segments $SM$ and $AD$ (see Fig. c). Then $L$ lies on the angle bisectors of angles $BAN$ and $ASN$, that is, it is the excenter of triangle $ANS$, and therefore lies on the angle bisector of angle $ANM$. On the other hand, from the equality of angles $SKM$ and $SMK$ of isosceles triangle $SKM$ and the parallelism of lines $AB$ and $CD$, it follows that $MK$ is the angle bisector of angle $NMC$. That is, $NL$ and $MK$ are the angle bisectors of the external angles of triangle $DMN$, and the point of their intersection is the excenter of this triangle, and therefore lies on diagonal $BD$ and coincides with point $P$. Then the desired angle is equal to angle $MPN$ (the angle between the angle bisectors of the external angles) and is equal to $45^{\circ}$.

In the case where point $P$ lies on segment $BO$, similar reasoning leads to the answer $135^{\circ}$.

## Answer

$45^{\circ}$ or $135^{\circ}$.