Prove that there exists a constant $δ ∈ [0,∞)$ such that the following holds. Let T be an arbitrary non-degenerate hyperbolic triangle in $\mathbb D$ with vertices $a, b, c ∈ \mathbb D$, and let $p_a ∈ [b, c]$,$ p_b ∈ [a, c]$ and $p_c ∈ [a, b]$ be such that
$d_\mathbb D(a, p_b) = d_\mathbb D(a, p_c)$, $d_\mathbb D(b, p_a) = d_\mathbb D(b, p_c)$ and $d_\mathbb D(c, p_a) = d_\mathbb D(c, p_b)$
Then for any points $q ∈ [a, p_c]$ and $r ∈ [a, p_b]$ with $d_\mathbb D(a, q) = d_\mathbb D(a, r)$ we have $d_\mathbb D(q, r) ≤ δ$
Conclude that any side of a hyperbolic triangle in the hyperbolic plane lies in the closed $δ$-neighborhood of the union of the two other sides.
Can I just say the following?
Let $δ = 2max[d_\mathbb D(a, p_b),d_\mathbb D(b, p_c),d_\mathbb D(c, p_a)]$.
As $d_\mathbb D(x, y)$ is defined to be the infimum of the length of all paths between $x$ and $y$, $d_\mathbb D(r, q) \le d_\mathbb D(a, r) + d_\mathbb D(a, q) \le d_\mathbb D(a, p_b) + d_\mathbb D(a, p_c) = 2d_\mathbb D(a, p_b) $.
You get similar results for the other 2 vertices and hence I choose $δ$ to be the maximum of $2d_\mathbb D(a, p_b), 2d_\mathbb D(b, p_c)$ and $ 2d_\mathbb D(c, p_a)$.
I am not sure at the moment if the below is true treat with care and be critical
OLD VERSION (no new one yet)
After some thinking I realised the answer is no. there is no such constant.
from $d(a, p_b) = d(a, p_c)$, $d(b, p_a) = d(b, p_c)$ and $d_(c, p_a) = d(c, p_b)$
you can calculate the distance $d(a,p_b) = d(a,p_c) = \frac{1}{2} \left( d(a,b)+ d(a,c)-d(b,c) \right) $ and has no maximum
Then Suppose $\angle a \ge $ right angle then $a$ is the closest point on $ ap_c$ to $ ap_b$ so $d(q,r) > d(a,r) $
Also suppose the point $r$ is the point $P_b$
The distance $d(a,r)$ has no maximum so also the distance $d(q,r)$ has no maximum
so there is no such constant.
I think you make the mistake that the shortest distance of q to any of the edges is to a point on $ab$ while even with the conditions you place on $q$ a point on $bc$ can be closer.
hope this helps