[Similar auxiliary triangles]  [Tangent properties and signs $]

Points $M$ and $N$ are the midpoints of the lateral sides $A C$ and $C B$ of the isosceles triangle $A C B$. Point $L$ is located on the median $B M$ such that

$B L: B M=4: 9$. The circle with center at point $L$ is tangent to the line $M N$ and intersects the line $A B$ at points $Q$ and $T$. Find the perimeter of triangle $M N C$, if $Q T=2, A B=8$.
Let $P$ be the point of tangency of the circle with the line $M N$, and $F$ be the projection of the center $L$ of the circle onto the line $A B$. Then the points $P, L$, and $F$ lie on the same line, and $F$ is the midpoint of $Q T$. Therefore, $F Q = F T = 1$. By the midline theorem of a triangle, $M N = \frac{1}{2} A B = 4$ and $M N \parallel A B$. Let $L P = L Q = L T = R$. Suppose point $Q$ lies between $A$ and $T$. From the similarity of triangles $L F B$ and $L P M$, we find that $L F = \frac{L B}{L M} \cdot L P = \frac{4R}{5}$. By the Pythagorean theorem, $L Q^2 = L F^2 + F Q^2$, or $R^2 = \frac{16}{25} R^2 + 1$, from which we get $R = \frac{5}{3}$, and $P F = L P + L F = \frac{9R}{5} = 3$.

Let $H$ be the projection of point $M$ onto the line $A B$. Then $A H = \frac{1}{2}(A B - M N) = 2$, and $M H = P F = 3$.

From the right triangle $A M H$, we find that $A M^2 = A H^2 + M H^2 = 13$. Therefore, $C N = C M = A M = \sqrt{13}$, and the perimeter of triangle $M N C$ is $\sqrt{13} + \sqrt{13} + 4 = 2(2 + \sqrt{13})$.

![](https://cdn.mathpix.com/cropped/2024_05_06_198acf192511df7a04ddg-27.jpg?height=1189&width=1455&top_left_y=0&top_left_x=342)

## Answer

$2(2 + \sqrt{13})$.