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

(N/A) Given: In $\Delta ABC$,$\angle C = 90^{\circ}$,$M$ is the mid-point of $AB$,and $MD \parallel BC$.
To prove: $CM = MA = \frac{1}{2} AB$.
Proof:
$1$. In $\Delta ABC$,since $MD \parallel BC$ and $M$ is the mid-point of $AB$,by the Converse of Mid-point Theorem,$D$ is the mid-point of $AC$. Thus,$AD = CD$.
$2$. Now,consider $\Delta ADM$ and $\Delta CDM$:
- $AD = CD$ (Proved above)
- $\angle ADM = \angle CDM = 90^{\circ}$ (Since $MD \parallel BC$ and $\angle ACB = 90^{\circ}$,$\angle ADM = \angle ACB = 90^{\circ}$ by corresponding angles)
- $DM = DM$ (Common side)
$3$. By $SAS$ congruence criterion,$\Delta ADM \cong \Delta CDM$.
$4$. By $c.p.c.t.$,$MA = MC$.
$5$. Since $M$ is the mid-point of $AB$,$MA = \frac{1}{2} AB$.
$6$. Therefore,$CM = MA = \frac{1}{2} AB$.