[  Transfer of side, diagonal, etc. ] Complexity [ Ratio of areas of triangles with a common base or common height ]

In trapezoid $A B C D$, the lateral side $A B$ is equal to the base $B C$, angle $B A D$ is $60^{\circ}$. Diagonal $B D$ is 3.

The area of triangle $A C D$ is to the area of triangle $A B C$ as $2: 1$. Find all sides of trapezoid $A B C D$.
Through vertex $C$, draw a line parallel to $AB$.

## Solution

Through vertex $C$, we draw a line parallel to the side $AB$, up to the intersection with the base $AD$ at point $K$. Then $ABCK$ is a rhombus.

Since

$$
\frac{S_{\triangle ACD}}{S_{\triangle ABC}}=\frac{AD}{BC}=2
$$

then $AD=2BC=2AK$. Therefore, $KBCD$ is also a rhombus with an acute angle of $60^{\circ}$.

If $M$ is the intersection point of the lines $CK$ and $BD$, then $BM$ is the height of the equilateral triangle $KBC$. Therefore,

$$
CM=BM \cot 60^{\circ}=\frac{3}{2} \cdot \frac{\sqrt{3}}{3}=\frac{\sqrt{3}}{2}
$$

Therefore,

$$
CD=BC=AB=AK=CK=\sqrt{3}, AD=2AK=2\sqrt{3}
$$

Answer

$$
AB=BC=CD=\sqrt{3}, AD=2\sqrt{3}
$$

In triangle $ABC$, angle $A$ is $45^{\circ}$, and angle $C$ is acute. A perpendicular $NM$ is dropped from the midpoint of side $BC$ to side $AC$. The areas of triangles $NMC$ and $ABC$ are in the ratio $1:8$. Find the angles of triangle $ABC$.

## Hint

$$
\frac{S_{\triangle NMC}}{S_{\triangle ABC}}=\frac{CN}{BC} \cdot \frac{CM}{AC}
$$

## Solution

Since

$$
\frac{S_{\triangle NMC}}{S_{\triangle ABC}}=\frac{CN}{BC} \cdot \frac{CM}{AC}=\frac{1}{2} \cdot \frac{CM}{AC}=\frac{1}{8}
$$

then $\frac{CM}{AC}=\frac{1}{4}$.

Let $BK$ be the height of triangle $ABC$. Then $NM$ is the midline of triangle $BKC$. Therefore, $KM=MC$ and $AK=KC$, i.e., triangle $ABC$ is isosceles. Therefore,

$$
\angle C=45^{\circ}, \angle B=90^{\circ}.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-46.jpg?height=363&width=763&top_left_y=1105&top_left_x=653)

Answer

$45^{\circ}, 90^{\circ}, 45^{\circ}$.