Question

$ABC$ is a triangle right angled at $C$ and $M$ is mid-point of hypotenuse $AB.$ Line drawn through $M$ and parallel to $BC$ intersects $AC$ at $D.$ then, $:$

$CM=MA=\frac{1}{2}AB$
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

Given : $△ABC$ is right angled at C, M is the mid-point of hypotenuse AB . $MD\parallel BC$
In $△ABC$, M is the mid-point of AB and $MD\parallel BC$
Hence by converse of mid point theorem, D is the mid-point of AC i.e. $AD=DC$
Since,$MD\parallel BC$ $\angle ADM=\angle ACB$ (Corresponding angles)
Hence, $\angle ADM={90}^{\circ }$
But, $\angle ADM+\angle CDM=90$ (Angles on a straight line)
$90+\angle CDM=180$
$\angle CDM={90}^{\circ }$
Thus, $MD\perp AC$
In $△AMD$ and $△CMD$,
$AD=CD$ (D is the mid-point of side AC)
$\angle ADM=\angle CDM$ (Each 90)
$DM=DM$ (Common)
$\therefore △AMD\cong △CMD$ (By SAS congruence rule)
Therefore, $AM=CM$ (By CPCT)
However, $AM=\frac{1}{2}AB$ (M is the mid-point of AB)
Hence,
$CM=AM=\frac{1}{2}AB$