Isosceles triangles $A B C(A B=B C)$ and $A_{1} B_{1} C_{1} \quad\left(A_{1} B_{1}=B_{1} C_{1}\right)$ are similar and $A C: A_{1} C_{1}=5: \sqrt{3}$.

Vertices $A_{1}$ and $B_{1}$ are located on sides $A C$ and $B C$, respectively, and vertex $C_{1}$ is on the extension of side

$A B$ beyond point $B$, and $A_{1} B_{1} \perp B C$. Find the angle $B$.
Prove that $C_{1} A_{1} \perp A C$ and formulate a trigonometric equation relative to the angle at the base of the isosceles triangle $A B C$.

## Solution

Let $A_{1} B_{1}=B_{1} C_{1}=1, \angle C=\gamma$. Then $A B=\frac{5}{\sqrt{3}}, \angle C_{1} A_{1} C=\angle C_{1} A_{1} B_{1}+\angle B_{1} A_{1} C=\gamma+\left(90^{\circ}-\gamma\right)=90^{\circ}, A_{1} C$ $=1 / \sin \gamma, C_{1} A_{1}=2 \cos \gamma, A A_{1}=C_{1} A_{1} \operatorname{ctg} \gamma=\frac{\cos ^{2} \gamma}{\sin \gamma}, A C=\frac{10}{\sqrt{3}} \cos \gamma$.

Since $A C=A A_{1}+A_{1} C$, then
$\sqrt{3} \operatorname{tg}^{2} \gamma-10 \operatorname{tg} \gamma+3 \sqrt{3}=0$

From this, $\operatorname{tg} \gamma=\frac{1}{\sqrt{3}}$ or $\operatorname{tg} \gamma=3 \sqrt{3}$.

The second solution does not satisfy the condition: point $C_{1}$ lies on the extension of segment $A B$ beyond point $B$, so

$A B \sin \gamma<A_{1} C_{1}$, that is, $\operatorname{tg} \gamma<\frac{2 \sqrt{3}}{5}$. Therefore, $\gamma=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_eab8c6bbe58aee0b55b3g-27.jpg?height=312&width=958&top_left_y=433&top_left_x=547)

Answer

$120^{\circ}$.

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