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The comparison triangle must have third side $d(x,y)$: the printed $d(y,z)$ contains an undefined $z$. Use $A=d(a,x)$, $B=d(a,y)$ and $C=d(x,y)$.

First extend the midpoint inequality along any constant-speed geodesic $\gamma$ from $p$ to $q$. If $L=d(p,q)$, the function $d(z,\gamma (t){)}^2-L^2{t}^2$ is midpoint-convex by (a). Iteration proves its convexity bound for dyadic $t$, and continuity then gives the Hadamard squared-distance inequalityApply this first to $[a,x]$ with $z=y_t$, and then to $[a,y]$ with $z=x$:Simplification givesIn a Euclidean comparison triangle, the cosine rule gives the scalar product of the two vectors from the common vertex as $(A^2+B^2-C^2)/2$. The squared distance between their multiples $s$ and $t$ is exactly the right side displayed above. Taking nonnegative square roots proves the required triangle comparison, including degenerate Euclidean triangles.

Let $p,q$ lie in the closed ball of center $z$ and radius $R$. On their unique geodesic, the Hadamard squared-distance inequality givesThus every point of that geodesic lies in the ball. By uniqueness there are no other geodesics to consider, so every closed ball is geodesically convex.