Prove that $A + B = C$ I drew the diagram here

I honestly do not see how $A$ and $B$ could possibly equal $C$.
 A: Hint: Observe that $0<A+B<\pi.$ What is $\tan(A+B)$? What is $\tan C$? Why is knowing these things enough?
A: $\sin A = \dfrac{1}{\sqrt{10}}$, $\cos A = \dfrac{3}{\sqrt{10}}$, $\sin B = \dfrac{1}{\sqrt5}$, $\cos B = \dfrac{2}{\sqrt5}$, $\sin C = \cos C = \dfrac1{\sqrt2}$.
$\sin(A + B)= \sin A \cos B + \cos A \sin B = \dfrac2{5\sqrt{2}} + \dfrac3{5\sqrt{2}} = \dfrac1{\sqrt{2}}= \sin C$
$\cos(A + B) = \cos A \cos B - \sin A \sin B = \dfrac6{5\sqrt{2}} - \dfrac1{5\sqrt{2}} = \dfrac1{\sqrt{2}} = \cos C$
Since their sines and cosines are equal, $C$ and $A + B$ are congruent modulo $2\pi$.  Since also $ 0 \lt A \lt B \lt C \lt \dfrac\pi{2}$, $-\dfrac{\pi}2 \lt C - (A + B) \lt \dfrac{\pi}2$, $A + B$ and $C$ must be equal.
A: Without trigonometry:

Green and yellow triangles are similar. Then, $\overline{s t}=2 \, \overline{r s}$ and $\overline{s t}^2+\overline{r s}^2=1$, therefore $\overline{r s}=\frac{1}{\sqrt{5}}$ and $\overline{s t}=\frac{2}{\sqrt{5}}$
Furthermore, $\overline{os}= \overline{r s}+ \sqrt{1^2+2^2}= \frac{6}{\sqrt{5}}$
Hence, $$\frac{\overline{o s}}{\overline{s t}} = 3$$
Then, identifying $\widehat{sot}=A$, we conclude that $B+A=C$
