# Q: 7       ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that               (iii)   $\small CM=MA=\frac{1}{2}AB$

Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.

To prove :  $\small CM=MA=\frac{1}{2}AB$

Proof : In $\triangle$ABC,

M is the midpoint of AB.                (Given)

DM || BC            (Given)

By converse of midpoint theorem,

D is the midpoint of AC  i.e.  AD = DC.

In $\triangle$ AMD  and $\triangle$ CMD,

$\angle$ADM = $\angle$ CDM    (Each  right angle)

DM = DM      (Common)

$\triangle$ AMD $\cong$$\triangle$ CMD      (By SAS)

AM = CM         (CPCT)

But ,$\small AM=\frac{1}{2}AB$

Hence,$\small CM=MA=\frac{1}{2}AB$

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