Question

$ABC$ is a triangle right angled at $C$. A line through the midpoint $M$ of hypotenuse $AB$ and Parallel to $BC$ intersects $AC$ at $D$. Show that
$CM=MA=\frac{1}{2}AB$

Answer 1

In $\Delta AMD$ And $\Delta ABC$

$\angle A$ is common;

$\angle B=\angle M$ (corresponding angles)

$\angle C=\angle D$ (BC is parallel to MD and BC is perpendicular to AC therefore, MD is also perpendicular to AC)..........(i)

$\Delta AMD\cong \Delta ABC$ (by AAA)

$\therefore \frac{AM}{AB}=\frac{AD}{AC}=\frac{1}{2}$

$\Rightarrow \frac{AD}{AD+CD}=\frac{1}{2}$

$\Rightarrow 2AD=AD+CD$

$\Rightarrow AD=CD$

Hence the D is mid point of AC

In $\Delta MDA$ and $△MDC$

$\angle MDA=\angle MDC={90}^0$

$DM=DM$

$CD=AD$ (proved above)

$\therefore \Delta MDA\cong \Delta MDC$......(SAS test)

$\therefore AM=MC=\frac{1}{2}AB$

Then M is the mid point of AB