10.389 In an isosceles triangle $A B C \quad A B=B C=25$ cm and $A C=14$ cm. Calculate the radius of the circle that touches $B C$ at point $D$, the foot of the altitude $A D$, and passes through the midpoint of $A C$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-237.jpg?height=333&width=317&top_left_y=947&top_left_x=148)

Let $O$ be the center of the circle passing through the midpoint $K$ of the base $AC$ and touching the side $BC$ at point $D$. Clearly, $BK = \sqrt{AB^2 - AK^2} = 24$.

$AC \cdot BK = BC \cdot AD \Rightarrow AD = 13.44 \Rightarrow$

$\Rightarrow \cos \angle CAD = \frac{AD}{AC} = 0.96$.

Let $x = OD = OK$, then by the cosine rule for

$\triangle AOK$, $x^2 = OK^2 = (AD - x)^2 + AK^2 -$

$- 2(AD - x)AK \cdot 0.96 \Rightarrow x = \frac{175}{48}$.

Answer: $\frac{175}{48}$ cm.