8.3. The height AH of isosceles triangle $\mathrm{ABC}$ with base $\mathrm{AC}$ is equal to the length of the perpendicular MK dropped from the midpoint M of side AB to the base of the triangle. Find the perimeter of triangle $\mathrm{ABC}$, if $\mathrm{AK}=\mathrm{a}$.
# 8.3. Answer: 20a.

Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_c73b449360b4321627c3g-1.jpg?height=551&width=354&top_left_y=1900&top_left_x=286)

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