Going through some Olympiad Math papers, I came across this question:
Given a rectangle $ABCD$ as shown in the figure below where $\overline{AB}=2$, $\overline{AD}=1$ and $M$ is on $\overline{CD}$. If $\overline{MA}$ bisects $\angle{DMB}$, find $\overline{DM}$.
This probably has something to do with trigonometry, but I'm suspecting that there's an easier way around this.
I set $x=\overline{DM}$ and tried:
$$2\arctan\left(\frac{1}{x}\right)=180-\arctan\left(\frac{1}{2-x}\right)$$
Because on the left side of the equation, it's the angle $\angle{DMA}$, and on the right side it's $\angle{DMB}$. This does not seem like it would be easy to solve, would there be an easier way around this?
Let $\angle DMA=\angle BMA=\alpha.$
Observe, $$\angle BAM=90^{\circ}-\angle DAM=\angle DMA=\alpha.$$ $$\implies BM=BA=2.$$ Using the Pythagorean Theorem, $CM=\sqrt{3}\implies DM=2-\sqrt{3}.$