In right triangle $ABC$,right-angled at $C$,$M$ is the mid-point of hypotenuse $AB$. $C$ is joined to $M$ and produced to a point $D$ such that $DM = CM$. Point $D$ is joined to point $B$ (see Fig). Show that:
$(i)$ $\Delta AMC \cong \Delta BMD$
$(ii)$ $\angle DBC$ is a right angle.
$(iii)$ $\Delta DBC \cong \Delta ACB$
$(iv)$ $CM = \frac{1}{2} AB$

(N/A) Given: In $\Delta ABC$,$\angle C = 90^\circ$ and $M$ is the mid-point of $AB$ $(AM = BM)$. $CM$ is produced to $D$ such that $DM = CM$.
$(i)$ In $\Delta AMC$ and $\Delta BMD$:
$AM = BM$ (Given,$M$ is the mid-point of $AB$)
$\angle AMC = \angle BMD$ (Vertically opposite angles)
$CM = DM$ (Given)
Therefore,by $SAS$ congruence criterion,$\Delta AMC \cong \Delta BMD$.
$(ii)$ Since $\Delta AMC \cong \Delta BMD$,their corresponding parts are equal $(c.p.c.t.)$.
So,$\angle MAC = \angle MBD$. These are alternate interior angles,which implies $AC \parallel DB$.
Since $AC \parallel DB$ and $BC$ is a transversal,the sum of consecutive interior angles is $180^\circ$.
$\angle ACB + \angle DBC = 180^\circ$
$90^\circ + \angle DBC = 180^\circ$
$\angle DBC = 180^\circ - 90^\circ = 90^\circ$.
$(iii)$ In $\Delta DBC$ and $\Delta ACB$:
$DB = AC$ ($c.p.c.t.$ from $\Delta AMC \cong \Delta BMD$)
$\angle DBC = \angle ACB = 90^\circ$ (Proved above)
$BC = CB$ (Common side)
Therefore,by $SAS$ congruence criterion,$\Delta DBC \cong \Delta ACB$.
$(iv)$ Since $\Delta DBC \cong \Delta ACB$,their corresponding parts are equal $(c.p.c.t.)$.
So,$DC = AB$.
Since $DM = CM$,$CM = \frac{1}{2} DC$.
Substituting $DC = AB$,we get $CM = \frac{1}{2} AB$.