\left.\begin{array}{l}{[\text { Ratio of areas of triangles with a common angle }]} \\ {[\quad \text { Ratios of areas }}\end{array}\right]

On the sides $AB$, $BC$, and $AC$ of triangle $ABC$, points $C_{1}$, $A_{1}$, and $B_{1}$ are taken respectively, such that

$$
\frac{A C_{1}}{C_{1} B}=\frac{B A_{1}}{A_{1} C}=\frac{C B_{1}}{B_{1} A}=2
$$

Find the area of triangle $A_{1} B_{1} C_{1}$, if the area of triangle $ABC$ is 1.
Triangles cut off from a given one by the lines $A_{1} B_{1}, B_{1} C_{1}$, and $A_{1} C_{1}$ are equal in area.

## Solution

Notice that

$$
s_{\Delta \mathrm{C}_{1} \mathrm{BA}_{1}}=\frac{B C_{1}}{B A} \cdot \frac{B A_{1}}{B C} \cdot s_{\Delta \mathrm{ABC}}=\frac{2}{9}.
$$

Similarly,

$$
S_{\Delta \mathrm{A}_{1} \mathrm{BC}_{1}}=S_{\Delta \mathrm{B}_{1} \mathrm{CA}_{1}}=\frac{2}{9}.
$$

Therefore,

$$
S_{\Delta \mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1}}=1-3 \cdot \frac{2}{9}=\frac{1}{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_02603d12f3437ea52c37g-22.jpg?height=391&width=698&top_left_y=2177&top_left_x=683)

## Answer

From the midpoint of the base of a triangle, lines parallel to the lateral sides are drawn. Prove that the area of the parallelogram formed in this way is half the area of the triangle.

## Hint

The area of a parallelogram is equal to the product of its adjacent sides and the sine of the angle between them.

## Solution

Let $M$ be the midpoint of the base $A B$ of triangle $A B C ; P$ and $Q$ be the vertices of parallelogram $M P C Q$, lying on sides $A C$ and $B C$. Then $P$ and $Q$ are the midpoints of sides $A C$ and $B C$. Therefore,

$$
S_{\mathrm{MPCQ}}=C Q \cdot C P \sin \angle C=\frac{1}{2} B C \cdot \frac{1}{2} A C \sin \angle C=\frac{1}{2} S_{\triangle \mathrm{ABC}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_02603d12f3437ea52c37g-23.jpg?height=395&width=824&top_left_y=703&top_left_x=617)

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