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**Hint:**In order to solve this question we have to prove that $\vartriangle DBC$ and $\vartriangle ACB$ are congruent at the very first instance and then by applying the midpoint concept the question can be solved.

**Complete step-by-step answer:**

From the $\Delta AMC$ and $\vartriangle BMD$ we get-

$AM = BM$ since $M$ is the midpoint of $AB$

$\angle AMC$$ = \angle BMD$

$CM = DM$ already given in the question

$\vartriangle AMC \cong \vartriangle BMD$ as it satisfy the condition of congruency (S-A-S)

Therefore, $AC = BD$ since it is corresponding parts of corresponding triangles.

Now in $\Delta DBC$ and $\Delta ACB$ we get-

$DB = AC$ which is already proved

$\angle DBC = \angle ACB$since both are ${90^0}$

$BC = CB$ as both are same

Therefore $\vartriangle DBC \cong \vartriangle ACB$ as it satisfies the condition of congruency

$AB = DC$ already proved

$AB = 2CM$ since $M$ is the midpoint

$CM = \dfrac{1}{2}AB$

Hence Proved.

**Note:**When all the three sides and three angles of the two triangles are equal then the two triangles can be said to be congruent.

The main conditions for congruency are given below:

If all the sides of one triangle are equal to all the three sides of the other triangle, then the two triangles are said to be congruent.

If two sides of a triangle and the angle present between them is equal to the other two sides and angle present between them of another triangle, then the two triangles are regarded as congruent.

If any two angles and a side of a triangle is equal to the two angles and one side of the other triangle, then the two triangles are said to be congruent.

If the hypotenuse and one side of a right- angled triangle is equal to the hypotenuse and a side of the second right- angled triangle, then the two right triangles are said to be congruent by RHS rule.

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