[Similar auxiliary triangles]

$[\quad$ Tetrahedron (other). $]$

A plane passes through the midpoints of edges $A B$ and $A C$ of the pyramid $A B C D$ and divides edge $B D$ in the ratio $1: 3$. In what ratio does this plane divide edge $C D$?
Let $M$ and $N$ be the midpoints of edges $AB$ and $AC$ respectively, and $K$ be a point on edge $BD$ such that $BK:KD = 1:3$. The line $MN$ is parallel to edge $BC$ (by the theorem of the midline of a triangle), so the line $MN$ is parallel to the plane $BCD$. The intersecting plane passes through the line $MN$, which is parallel to the plane $BCD$ and has a common point $K$ with this plane. Therefore, the intersecting plane intersects the plane $BCD$ along a line $l$ parallel to $MN$, and hence to $BC$. Let the line $l$ intersect edge $CD$ at point $L$. Then $L$ is the point of intersection of the intersecting plane with edge $CD$. Since $KL \parallel BC$, we have

$CL:LD = BK:KD = 1:3$.

## Answer

In the convex quadrilateral $ABCD$, the equalities $\angle CBD = \angle CAB$ and $\angle ACD = \angle ADB$ hold.

Prove that a right triangle can be formed from the segments $BC$, $AD$, and $AC$.

## Solution

Let the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $O$. Triangles $ABC$ and $BOC$ are similar by two angles. Therefore,

$BC:OC = AC:BC$, from which $BC^2 = AC \cdot OC$. Similarly, $AD^2 = AC \cdot AO$ and $BC^2 + AD^2 = AC \cdot OC + AC \cdot AO = AC^2$.

Thus, by the converse of the Pythagorean theorem, a right triangle can be formed from the segments $BC$, $AD$, and $AC$.

![](https://cdn.mathpix.com/cropped/2024_05_06_198acf192511df7a04ddg-20.jpg?height=200&width=228&top_left_y=612&top_left_x=915)

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