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
$\left(i\right)D$ is the mid-point of $AC$
$\left(ii\right)MD\perp AC$
$\left(iii\right)CM=MA=\frac{1}{2}AB$

Answer 1

REF.Image

Given : $AM=MB$ & $MD\parallel BC$

Since $MD\parallel BC$

$\angle ADM=\angle ACB$ (Corresponding angles)

$\angle ADM={90}^{\circ }$

Hence , $MD\perp AC$

Proved.

Since $DM\parallel BC$, By Basic proportionality Theorem

$\left(\frac{AD}{AC}\right)=\left(\frac{AM}{AB}\right)=\frac{1}{2}$

$\Rightarrow AC=2AD$

Thus $D$ is mid point of $AC$ Proved!

In $\Delta ADM$ & $\Delta CDM$

$AD=CD$ (proved above)

$\angle ADM=\angle CDM$ $(={90}^{\circ })$

$DM=DM$ (common side)

$\therefore \Delta ADM\cong \Delta CDM$ by SAS

Hence by cpct

$AM=CM$

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

$\therefore AM=CM=\frac{1}{2}\left(AB\right)$