[ Pythagorean Theorem (direct and inverse).]

In triangle $ABC$, it is known that $AB=3$, height $CD=\sqrt{3}$. The base $D$ of height $CD$ lies on side $AB$ and $AD=BC$. Find $AC$.
Let $B C = A D = a$. By the Pythagorean theorem, $a^{2} = (3-a)^{2} + 3$, from which $a = 2$. Therefore, $A C^{2} = 4 + 3 = 7$.

![](https://cdn.mathpix.com/cropped/2024_05_06_04ea0aec75eb28164af1g-21.jpg?height=343&width=826&top_left_y=980&top_left_x=616)

Answer

$\sqrt{7}$.