On the extension of the bisector $A L$ of triangle $A B C$ beyond point $A$, a point $D$ is taken such that $A D=10$ and $\angle B D C = \angle B A L = 60^{\circ}$.

Find the area of triangle $A B C$. What is the smallest area of triangle $B D C$ under these conditions?
Prove the similarity of triangles $A B D$ and $A D C$. To find the minimum area, use the Cauchy inequality.

## Solution

By the external angle theorem of a triangle, $\angle A B D=60^{\circ}-\angle B D A=\angle A D C$.

Thus, triangles $A B D$ and $A D C$ are similar by two angles (their angles at the common vertex $A$ are equal to $120^{\circ}$), so $A B: A D=A D: A C$. Therefore,

$A B \cdot A C=A D^{2}=100$.

Hence, $S_{A B C}=1 / 2 A B \cdot A C \sin \angle A=25 \sqrt{3}$.

$$
S_{A D C}=S_{A B D}+S_{A D C}+S_{A B C}=1 / 2 A D \cdot A B \sin 120^{\circ}+1 / 2 A D \cdot A C \sin 120^{\circ}+25 \sqrt{3}=5 A B \cdot \frac{\sqrt{3}}{2}+5 A C \cdot \frac{\sqrt{3}}{2}+25 \sqrt{3}
$$

$\geq 5 \sqrt{3}(\sqrt{A B \cdot A C}+5)=75 \sqrt{3}$, and equality is achieved if $A B=A C=10$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-34.jpg?height=595&width=724&top_left_y=0&top_left_x=667)

## Answer

$25 \sqrt{3}, 75 \sqrt{3}$