On pg. 400 of the book Metric Spaces of Non-Positive Curvature by Bridson and Haefliger, it is mentioned that:
The hyperbolic plane $\mathbf H^2=\{(a, b)\in \mathbf R^2:\ b>0\}$, (with Riemannian metric $(dx^2+dy^2)/y^2$) is $\delta$-hyperbolic for some $\delta>0$.
(A complete Riemannian manifold is said to be $\delta$-hyperbolic if each side of a geodesic triangle is contained in the union of the $\delta$-neighborhood of the other two. If such a delta exists, we say the manifold is hyperbolic.)
Towards the proof of the above mentioned statement, the authors write
To see that $\mathbf H^2$ is hyperbolic, note that since the area of geodesic triangles in $\mathbf H^2$ is bounded by $\pi$, there is a bound on the radius of semi circles that can be inscribed in a geodesic triangle.
I can see that the area of any goedesic triangle is bounded by $\pi$. This is immediate by applying the Gauss-Bonnet formula and using the fact that $\mathbf H^2$ has constant curvature $-1$, since the sum of exterior angles of any triangle cannot exceed $3\pi$.
But I am unable to see how this gives us $\delta$-hyperbolicity of $\mathbf H^2$. I do not follow what the authors mean by "inscribing a semi circle in a geodesic triangle."
The definition of $\delta$-hyperbolic space says: any point on a side of a geodesic triangle is within the distance $\le \delta$ of some other side.
If the above fails, then some side (say, $AB$) of some geodesic triangle contains a point $P$ at large distance $R$ from other two sides $BC, AC$. Draw the circle of radius $R$ (in the hyperbolic metric) centered at $P$. The side $AB$ divides it evenly in two halves -- it's a diameter of the circle, since it is a geodesic passing through the center. So, one half of the circle is within the triangle. This implies that the triangle has large area, a contradiction.