Value of the 'height' of a triangle

Question

Given the length of $AB = x_1 = 10$ and $AC = x_2 = 14$, how do I find out the length of $AD = x_3$ and $BC$?

Is $AD = \frac{1}{2}(x_1+x_2)$

and

$BC = x_2-x_1$?

Answer 1

There seems to be some information missing, but if we make some assumptions, we may be able to generalize to where the assumptions don't matter.

Now, it looks like $\triangle ABC$ is a right triangle with hypotenuse $x_2$. It also appears that angle $A$ is 45º. If these two assumptions are correct, then we have an isosceles right triangle. The measurements of $x_1=10$ and $x_2=14$ provide validity that these assumptions are correct. (WHY? I leave this as an exercise for the OP.)

A third assumption we can make is that $x_3$ bisects side $BC$. If this is true, then $x_3$ bisects the angle $A$ (which we are assuming to be 45º). Using $SOH-CAH-TOA$ then, we have $\cos22.5 =\frac{x_1}{x_3}$, so $x_3\approx10.8$.

If we discard our assumptions and find more data regarding the triangle, then we have, in general, $x_3=x_1\div\cos(BAD)$.