294. The radius (4) of the circle inscribed in a triangle and the segments 6 and 8, into which the point of tangency divides one side of the triangle, are given. Find the other two sides.
294. It is clear (Fig. 66) that $A D=6$ and $E C=8$; let $B D=$ $=B E=x$. On one hand,

![](https://cdn.mathpix.com/cropped/2024_05_21_926f0bb262c8f9569516g-143.jpg?height=394&width=574&top_left_y=1528&top_left_x=364)

Fig. 66. The area of the triangle:

$$
\begin{aligned}
\Delta & =A O C+A O B+B O C= \\
& =[14 \cdot 4+(6+x) 4+ \\
+ & (8+x) 4] \frac{1}{2}=56+4 x
\end{aligned}
$$

The perimeter

$2 p=2 x+12+16 ; p=x+14 ;$ therefore,

from which

$$
\Delta=\sqrt{(x+14) \times 6 \cdot 8}
$$

$14+x=\sqrt{3 x(x+14)} ;(14+x)^{2}=3 x(x+14) ; 14+x=3 x$ and

$$
x=7
$$

therefore, $A B=13 ; B C=15$.

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