Given a right triangle $\,(A,B,C)\,$ with legs $x$ and $y,\,$ there is a point $h$ on the hypotenuse dividing the latter into $h_1$ (from the origin to $h$), and $h_2$ (from $h$ to the end of the hypotenuse away from the origin.
The Distance $\,D\,$ from the right angle to a point $\,h\,$ can be found using the fact that the triangle with vertices (x,h, origin) is similar to $(A,B,C)$ and can be found using the trig formula $\,D=A\sin\theta=A\dfrac{B}{C}=\dfrac{AB}{C}.\quad$ For example, in a $(3,4,5)$ triangle, $D=\dfrac{3\cdot4}{5}=2.4.$
I was wondering if there was a way to find this distance without using trig, using just the Pythagorean Theorem. I have stated with these equations but I don't know where to go from here.
\begin{align*} H_1^2+D^2&=A^2\\ H_2^2+D^2&=B^2 \\ D^2&=A^2-H_1^2 \\ D^2&=B^2-H_2^2\\ \\ \implies A^2-H_2^1&=B^2-H_2^2\\ \\ \implies A^2-B^2&=H_1^2-H_2^2 \end{align*}
We might have something if we could arrange these to find $H_1+H_2$ but that doesn't seem likely.
You absolutely can do it. Here's one way, working from what you've already done as a starting point but also adding two more facts: $C = H_1 + H_2$ and $C^2 = A^2 + B^2$.
So let's start with the point you go to, with $A^2 - B^2 = H_1^2 - H_2^2$. We can factor the RHS as $H_1^2 - H_2^2 = (H_1 - H_2)(H_1 + H_2) = (H_1 - H_2)C$. Then we can perform some linear elimination to get:
$\begin{eqnarray} & C(H_1 - H_2) & = & A^2 - B^2 \\ & H_1 - H_2 & = & \frac{1}{C}(A^2 - B^2) & (1) \\ & H_1 + H_2 & = & C & (2) \\ (1) + (2): & 2 H_1 & = & C + \frac{1}{C}(A^2 - B^2) \\ && = & \frac{1}{C}(C^2 + A^2 - B^2) \\ && = & \frac{1}{C}((A^2 + B^2) + A^2 - B^2) \\ && = & \frac{2A^2}{C} \\ & \frac{H_1}{A} & = & \frac{A}{C} \end{eqnarray}$
And that gives us the similarity relation that you can then plug into $A^2 + H_1^2 = D^2$ to get the desired final result.