If we have two triangles $akz$ and $bkc$ in which they have same side $k$.
If $t_1$ and $t_2$ are two angles $\in \thinspace]0\thinspace , \thinspace 90[$ in which the above-mentioned $k$ is the adjacent as shown in the following image:
Can we prove that:
if $b > a > k$ $\Rightarrow$ $c > z$ is always true?
No. $X = (0, 0)$, $Y = (4, 0)$ and $Z = (4, 2)$, and $W = (5, 1)$. Here $k = [X, Y]$, $a = [X, Z]$, $ z = [Y, Z]$ and $ b = [X, W]$ and $ c = [Y, W]$. Note, $|k| = 4$, $|a| = \sqrt{20}$ $|z| = 2$, $|b| = \sqrt{26}$, and $|c| = \sqrt{2}$.
Geometrywise, $b$ is outside the circle of radius $|a|$ and centre $X$ (not too far) and much nearer the line $k$.