8.3. In an isosceles triangle \(ABC (AB = BC)\), the altitude \(AH\) is drawn, and from the midpoint \(M\) of side \(AB\), a perpendicular \(MK\) is dropped to side \(AC\). It turns out that \(AH = MK\). Find the perimeter of triangle \(ABC\) if \(AK = a\).
Answer: 20a.

Solution. Triangle $ABC$ is isosceles, so $\angle MAK = \angle ACH$. Therefore, triangles $MAK$ and $ACH$ are congruent (right triangles with equal legs $MK = AH$ and equal acute angles opposite these legs). This means $AC = AM = \frac{1}{2} AB$. Draw the altitude $BO$ of triangle $ABC$. By Thales' theorem, $AK: KO = AM: MB$, so $AO = 2AK = 2a$. Then $AC = 2AO = 4a$, and $AB = 2AC = 8a$. Therefore, the perimeter is $8a + 8a + 4a = 20a$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3fedebe1ade412f64b35g-2.jpg?height=728&width=508&top_left_y=221&top_left_x=914)

Comment. The congruence of triangles $MAK$ and $ACH$ has been proven - 2 points.