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. Then, mark the correct answer option.

• A. $CM=CD$
• B. $CM=MA=\frac{1}{2}AB$
• C. $CM=MA=\frac{1}{2}AC$
• D. $Noneofthese$

Answer 1

The correct option is B $CM=MA=\frac{1}{2}AB$




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.