$\left.\begin{array}{l}{\left[\begin{array}{l}\text { Two pairs of similar triangles } \\ \text { [ }\end{array}\right]}\end{array}\right]$

Given trapezoid $A B C D$. A line parallel to its bases intersects the lateral sides $A B$ and $C D$ at points $P$ and $Q$, respectively, and the diagonals $A C$ and $B D$ at points $L$ and $R$, respectively. The diagonals $A C$ and $B D$ intersect at point $O$. It is known that $B C=a, A D=b$, and the areas of triangles $B O C$ and $L O R$ are equal. Find $P Q$, if point $L$ lies between points $A$ and $O$.
Prove that ${ }^{A L} /{ }_{A C}={ }^{b-a /{ }_{b}+a}$.

## Solution

From the condition, it follows that $b>a$. Triangles $B O C$ and $R O L$ are similar, and since they are equal in area, they are equal. From the similarity of triangles $B O C$ and $D O A$, it follows that $A O: O C=b: a$.

Therefore, ${ }^{A L} / L C=b-a / 2 a,{ }^{A L} / A C=b-a / b+a$.

From the similarity of triangles $A P L$ and $A B C$, it follows that $P L: B C=A L: A C$.

Thus, $P L=a(b-a) / b+a$.

Similarly, $R Q=a(b-a) / b+a$. Therefore, $P Q=P L+L R+R Q=2 a(b-a) / b+a+a=a(3 b-a) / b+a$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5ddd4eab4fe1a12edeb9g-01.jpg?height=315&width=554&top_left_y=1749&top_left_x=755)

Answer

$a(3 b-a) / b+a \cdot$