 Solution

Work on the unit sphere, so side lengths are central angles. Place the right-angle vertex at the north pole and choose perpendicular meridians through it. The other vertices can then be writtenTheir dot product is the cosine of the central angle between them, namely the hypotenuse length $c$. Hence the spherical Pythagorean relation isFor the small triangle let its hypotenuse be $c_{\lambda }$. The same formula gives $\cos c_{\lambda }=\cos (\lambda a)\cos (\lambda b)$, so $c_{\lambda }\to 0$. Expanding each cosine about zero yieldsIn particular $c_{\lambda }=O(\lambda )$, and thus $c_{\lambda }^2={\lambda }^2(a^2+b^2)+O({\lambda }^4)$. Consequently $(c_{\lambda }/\lambda {)}^2\to a^2+b^2$, the ordinary Pythagorean theorem in the locally flat limit.