Does exist a triangle with sides a integer length where one of height is equal to the side which is the base? Does exist a triangle with sides a integer length where one of height is equal to the side which is the base?

$a,k,l$ -natural,
$a$ is lenght of hight

[Note: This is probably the same as what Dietrich said, only I like to see the geometry.]
The question is equivalent to asking if there is a triangle with rational side lengths, since we can just scale by the product of the denominators and get integers. So we can assume that $a=1$, since we can just scale by the inverse of whatever rational number $a$ is.
Geometrically, this allows us to put our triangle in the $1\times 1$ square in the cooridinate plane.  This now only works for triangles with the third vertex between (0,1) and (1,1). I suppose we would need to modify this to use the law of cosines for the rest of the triangles. 
Then it is just the pythagorean theorme and we have that $$\ell^2 =x^2+1$$ and $$k^2=(1-x)^2+1.  $$
We want $k$ and $\ell$ to be rational, so lets just call $\ell=\frac{p}{q}$.  Then $$x^2=\frac{p^2}{q^2}-1= \frac{p^2-q^2}{q^2},$$ thus
$$x= \frac{\sqrt{p^2-q^2}}{q}.$$
So with some replacement and simplification, $$k^2=2-2x+x^2$$
$$k^2=2-2\frac{\sqrt{p^2-q^2}}{q}+\frac{p^2-q^2}{q^2}$$
$$k^2=\frac{2q^2}{q^2}-\frac{2q\sqrt{p^2-q^2}}{q^2}+\frac{p^2-q^2}{q^2}$$
$$k^2=\frac{q^2-2q\sqrt{p^2-q^2}+p^2}{q^2}$$
$$k=\frac{\sqrt{q^2+p^2-2q\sqrt{p^2-q^2}}}{q}$$
Then all we need is that the numerator is an integer.  I am in the process of making mathematica find one, but I checked the first $1000\times 1000$ pairs of integers for $p<q$ and came up with nothing...  Will let you know if I find something though.
Thanks to TonyK for finding a mistake of mine.
