Prove the excircle and right triangle inequality 
Let $\triangle{ABC}$ be a right triangle with $\angle{A}= 90^{\circ}$. Let $D$ be the intersection of the internal angle bisector of $\angle{A}$ with side $BC$ and $I_a$ be the center of the excircle of the triangle ABC opposite to the vertex $A$. Prove that $$\dfrac{DA}{DI_a} \leq \sqrt{2}-1.$$

I tried using reflections since $\angle{BI_aC} = 45^{\circ}$ we know that $I_a$ lies on a circle with $A$ as the center. I couldn't really find out how to incorporate the ratio. In case anyone wants a diagram to see this I drew one (ignore the scratch work to the side of it).


Let circle O be the excircle (so O is its center). It must be tangent to its intersection with lines AB and AC by definition (call the intersection points X and Y). Then angles AXO and AYO are right angles, and by how X and Y were defined, so is XAY. Since XO and YO are radii of circle O, this means XO=YO, and thus quadridlateral XAYO is actually a square.
You also know that XO and YO are the same length as EO (where E is the point of tangency on BC), since EO is also a radius of circle O. Then you can show that equality holds in the problem statement when where AB=AC, and the ratio only gets smaller (or stays the same) if AB $\neq$ AC
