255. In a triangle, the difference between two lateral sides is 3, the height is 10, and the difference between the segments of the base formed by the foot of the height is 12. Find the sides.
255. Let $A C - A B = F C = 3 ; \quad A D = 10 ; \quad D C - B D = E C = 6 ; \quad B D = \frac{B C - 6}{2} = \frac{B C}{2} - 3$ (Fig. 58).

If $B C = x$, then $B D = \frac{x}{2} - 3$. Further,

$$
A C - A B = 3 = \frac{6}{2} = \frac{1}{2}(D C - B D)
$$

$A D^{2} = A C^{2} - D C^{2} = A B^{2} - B D^{2}$, from which $A C^{2} - A B^{2} = D C^{2} - B D^{2}$, or $\quad (A C - A B)(A C + A B) = (D C - B D)(D C + B D)$. 134

Dividing by $A C - A B$, we get:

$$
A C + A B = 2 x ; A B = 2 x - A C = 2 x - A B - 3
$$

or

$2 A B = 2 x - 3 ; A B = x - \frac{3}{2}$.

Now

$A B^{2} = x^{2} - 3 x + 2 \frac{1}{4} ; B D^{2} = \frac{x^{2}}{4} - 3 x + 9$.

![](https://cdn.mathpix.com/cropped/2024_05_21_926f0bb262c8f9569516g-136.jpg?height=271&width=460&top_left_y=390&top_left_x=1249)

Fig. 58.

But $A D^{2} + B D^{2} = 100 + \frac{x^{2}}{4} - 3 x + 9 = A B^{2} = x^{2} - 3 x + 2 \frac{1}{4}$

from which $100 + \frac{27}{4} = x^{2} - \frac{x^{2}}{4} ; 3 x^{2} = 427$;

$$
x = \sqrt{\frac{427}{3}}
$$

Now it is easy to find $A B$ and $A C$.