Question

$ABC$ 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
$CM=MA=\frac{1}{2}AB$

Answer 1

Consider a right angle $\Delta ABC$ right angled at C.

$\Rightarrow \angle C={90}^{\circ }$

And $M$ is the mid-point of $AB.$ and DM $\parallel$ BC

$\Rightarrow D$ is the midpoint of AC [By converse of midpoint theorem]

In $\Delta ADM$ and $\Delta CDM$ ,

$DM=MD$ [Common]

$AD=CD$ [Since, D is mid-point of AC]

$\angle ADM=\angle CDM$ [Since $DM$ parallel BC, $\angle$ ADM = ${90}^{\circ }$]

$\Delta ADM\cong \Delta CDM$ [SAS congruency]

$CM=AM$……(i) [CPCT]

$AM=BM=\frac{1}{2}AB$..........(ii)

From Equations (i) and (ii) , we get

$CM=AM=\frac{1}{2}AB$

Hence , it is proved.