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

Hint: Vertically opposite angles are equal and sum of angles in a triangle is equal to 180.
From triangle ABC, we have a perpendicular angle at C. For the first part, we use SAS rule to prove congruence rule to prove the congruence. When we are proving $\angle DBC$ is right angled, we use the sum of angles in the triangle as ${180^o}$.

Complete step by step solution:
First, we can see that in $\vartriangle ABC,\angle C = {90^o}$
i)Now, proving the congruence of the triangles $\vartriangle DMB,\vartriangle AMC$
We are given that $DM = MC$
Since M is the midpoint of AB, we can say that
$AM = MC$
If we consider AB and CD are intersecting, then vertically opposite angles are equal.
$\angle DMC = \angle AMC$
Therefore ,we have proved $\vartriangle DMB \cong \vartriangle AMC$

ii) Now, let's prove that $\angle DBC$ is a right angle.
We can see that MC is bisecting the $\angle C$ in the \[\vartriangle DBC \cong \vartriangle ACB\].
Therefore
$\angle MCB = \angle MCA = {45^o}$.
Since, we know that alternate angles are equal.
\[\angle MCA = \angle DCB = {45^o}\]
Now, consider \[\vartriangle DBC\]
The sum of angles in a triangle is 180.
\[{45^o} + {45^o} + \angle DBC = {180^o}\]
\[\angle DBC = {90^o}\]
From, that we can say that
\[\angle DBC\] is right angle triangle
iii) Now, we are going to prove \[\vartriangle DBC \cong \vartriangle ACB\]
BC is the common side to the two triangles.
$\angle ACB = \angle DBC = {90^o}$
Since DB and AC are parts of congruent triangles. They are proved equal due to CPCT ( corresponding parts to corresponding triangles).
Therefore
\[
MC = \dfrac{1}{2}(AB) \\
   = \dfrac{1}{2}(AB) \\
\]by SAS rule.
Also, we can conclude that
\[AB = {\text{ }}CD\] by CPCT.
As, we know that \[DM{\text{ }} = {\text{ }}MC\]
We can say that
\[
  MC = \dfrac{1}{2}(CD) \\
   = \dfrac{1}{2}(AB) \\
 \]
Therefore
\[MC = \dfrac{1}{2}(AB)\]
Hence proved

Note: For CPCT to be applied, it is necessary that the side or anything, both the parts should be part of the triangles which have been proved congruent, or have been given as congruent triangles in the question. A bisecting line always divides angle into two equal angles.