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How can we find the upper bound of the hyperbolic distance from any point on the side $AB$ to either $AC$ and $BC$ for the hyperbolic triangle $\triangle ABC$? Help will be appreciated.

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I get $\log (1 + \sqrt 2).$ Take the upper half plane, with the triply infinite triangle with edges $$ x=-1, \; x^2 + y^2 = 1, \; x = 1, $$ all with $y >0.$ The point $x=0, \; y=1$ is at equal distance from the two vertical edges.

In this model, any non-vertical unit speed geodesic is given by $$ x = A + B \tanh t, \; \; y = B \; \mbox{sech} \; t $$ with $B > 0.$ The particular one that meets the edge $x=1$ orthogonally and passes through the point $x=0, y=1$ is $$ x = 1 + \sqrt 2 \; \tanh t, \; \; y = \sqrt 2 \; \mbox{sech} \; t. $$ The $t$ values at which this curve achieves $y=1$ satisfy $\cosh t = \sqrt 2,$ so $$ t = \pm \log (1 + \sqrt 2).$$ As the curve is on the edge $x=1$ when $t=0,$ the geodesic distance is $$ \log (1 + \sqrt 2). $$

You really ought to have said something about the minimum of the distances from a point on the edge to either of the other edges, then the maximum possible for that. All together, this is a minimax optimum.

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curiously gives another value of the edge length of the inscribed triangle. – Willemien Jan 7 at 21:25

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