$\left.\frac{\text { Auxiliary similar triangles }}{[\quad \text { Law of Cosines }}\right]$

In triangle $ABC$, a point $D$ is taken on side $AC$, such that $AD=3$, $\cos \angle BDC=13/20$, and $\angle B+\angle ADB=180^{\circ}$. Find the perimeter of triangle $ABC$ if $BC=2$.
Triangles $A B C$ and $B D C$ are similar.

## Solution

Since $\angle B=180^{\circ}-\angle A D B=\angle B D C$, triangles $A B C$ and $B D C$ are similar by two angles. Therefore, $B C$ : $A C=D C: B C$, or $\frac{2}{3+D C}=1 / 2 D C$. From this, we get $D C=1, A C=4$. By the cosine rule, $A C^{2}=A B^{2}+B C^{2}-$

$2 A B \cdot B C \cos \angle B$, or $16=A B^{2}+4-2 \cdot 2 A B \cdot 13 / 20$.

From this, we find that $A B=5$. Therefore, the perimeter of triangle $A B C$ is $A C+B C+A B=4+2+5=$ 11.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-23.jpg?height=286&width=460&top_left_y=1992&top_left_x=799)

## Answer

11.