What is the length of the shorter trisector of the right angle in a $3$-$4$-$5$ triangle?
I found this question in a local question paper, and I am unable to solve it. I applied Cosine formula, but I didn't got an elegant solution to it.
Any help will be appreciated.
$$\begin{align} |\triangle ABC| &\;=\; |\triangle BDC| + |\triangle ADC| \\[6pt] \implies \qquad \frac{1}{2}\,a b &\;=\; \frac{1}{2}\,a d \sin 30^\circ \;+\; \frac{1}{2}\,b d \sin 60^\circ \\[6pt] \implies \qquad \frac{1}{2}\,a b &\;=\; \frac{1}{2}\,a d\cdot\frac{1}{2} \;+\; \frac{1}{2}\,b d\cdot\frac{\sqrt{3}}{2} \\[8pt] \implies \qquad 2 a b &\;=\; d\left(\; a + b \sqrt{3} \;\right) \\[6pt] \implies \qquad d &\;=\; \frac{2a b}{a + b\sqrt{3}} \end{align}$$