For reference:
In the figure, calculate $MD$, if $BM = 6\ \mathrm m$, $DE= 1\ \mathrm m$ and $M$ is incenter *. (Answer:$2\ \mathrm m$)
*(Added later)
My progress: I found a solution on the net but the consideration of point $M$ being the incentre (not mentioned) was made in the statement besides using more advanced trigonometry. Would it be possible to solve it by geometry and without considering M as incenter?
Let $M$ = incenter triangle $ABC$
By Incenter Theorems:
$\displaystyle \frac{MB}{MD} = \frac{a+c}{b} \iff \frac{BD}{MD} = \frac{a+b+c}b \iff MD = \frac{b}{2p} \cdot BD = \frac b{2p} \cdot \frac{2ac \cdot \cos \frac{B}2}{a+c} \implies \\\displaystyle MB = \frac{ac \cos \frac{B}2}{p}\\ \displaystyle EM = EA = EC = \frac{\frac b2}{\cos \frac B2} = \frac b{2\cos \frac B2}$
Using a non-trivial relationship
$\displaystyle ac \cos^2 \frac B2 = p(p-b) \implies \\\displaystyle \frac{MB}{ME} = \frac{2(p-b)}b \iff \frac{6}{1 + MD} = \frac{a+c - b}{b} = \frac{MB}{MD} -1\\\displaystyle \implies \frac6{1+x} = \frac6x -1 \iff x=2$
Original Problem:
Say, $MD = x$. As $M$ is the incenter of $\triangle ABC$,
$ \displaystyle \frac{BM}{MD} = \frac{a+c}b$
$ \displaystyle \implies \frac 6x = \frac{a+c}b\tag1$
$ \displaystyle CD = \frac{ab}{a+c}\tag2$
As $\triangle ADE \sim \triangle BDC$ and $AE = ME$,
$ \displaystyle \frac{AE}{DE} = \frac{BC}{CD}$
$ \displaystyle \implies ME = MD + DE = x + 1 = \frac{a+c}b ~ \left[\text {using } (2)\right]$
Plugging into $(1)$,
$ \displaystyle \frac 6 x = x + 1 \implies (x-2)(x+3) = 0$
That leads to $x = 2$.
$AE = CE = ME$ is always true if $M$ is the incenter of $\triangle ABC$ and $E$ is the point where the angle bisector of $\angle B$ intersects the circumcircle of $\triangle ABC$. See below -
$ \displaystyle \angle CAE = \frac{\angle B} 2, \angle MAC = \frac{\angle A}2$
$ \displaystyle \implies \angle MAE = \frac{\angle B + \angle A}2 = 90^\circ - \frac{\angle C}2$
Also, $ \displaystyle \angle MEA = \angle C \implies \angle AME = 90^\circ - \frac{\angle C}2 = \angle MAE$
$\therefore AE = ME$