$ABC$ is a triangle right angled at $C$. A line through the mid-point $M$ of hypotenuse $AB$ and parallel to $BC$ intersects $AC$ and $D$. Show that
(i) $D$ is the mid-point of $AC$
(ii) $MB\perp AC$
(iii) $CM=MA=\frac{1}{2}AB$

(i) In $△ABC$, 

$M$ is the mid point  of $AB$ and $MD\parallel BC$. 

Therefore, $D$ is the mid- point of $AC$. 
i.e., $AD=DC$      ... (1)

(ii) Since $MD\parallel BC$, corresponding angles has to be equal

$\angle ADM=\angle ACB$

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

As $\angle ACB=90^{\circ }$  is given

Since $\angle ADM$ and $\angle CDM$ are angles of a linear pair

But $\angle ADM+\angle CDM=180^{\circ }$

Therfore $90^{\circ }+\angle CDM=180^{\circ }\Rightarrow \angle CDM=90^{\circ }$

Thus, $\angle ADM=\angle CMD=90^o$    ... (2)  

$\Rightarrow MD\perp AC$