In a triangle, the sides are known: $A B=15, B C=13$ and $A C=14$. Through point $C$, a perpendicular is drawn to side $A C$ until it intersects at point $K$ with the extension of side $A B$. Find $B K$ and $C K$.

#
Find the height $BP$ of triangle $ABC$.

## Solution

Note that triangle $ABC$ is acute-angled. Let $P$ be the projection of vertex $B$ onto side $AC$. Denote $CP=x$. Then $AP=14-x$, and since

$$
BC^2 - CP^2 = AB^2 - AP^2
$$

$x$ is the root of the equation

$$
196 - x^2 = 225 - (14 - x)^2
$$

i.e., $x=5$.

Let $\angle CBP = \alpha$, $\angle ABP = \beta$. Then

$$
\sin \angle KCB = \sin \alpha = \frac{CP}{BC} = \frac{5}{13}, \sin \angle CPB = \sin \beta = \frac{AP}{AB} = \frac{9}{15} = \frac{3}{5}
$$

From triangle $CBK$, by the Law of Sines, we find that

$$
BK = \frac{BC \sin \alpha}{\sin \beta} = \frac{25}{3}
$$

By the Pythagorean theorem,

$$
CK = \sqrt{(AB + BK)^2 - AC^2} = \frac{56}{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-29.jpg?height=385&width=820&top_left_y=-1&top_left_x=616)

## Answer

$\frac{25}{3}$ and $\frac{56}{3}$.