In a right triangle,prove that the line segment joining the mid-point of the hypotenuse to the opposite vertex is half the length of the hypotenuse.

(N/A) $ABC$ is a right triangle,right-angled at $B$,and $D$ is the mid-point of the hypotenuse $AC$. We need to prove that $BD = \frac{1}{2} AC$.
Construction: Produce $BD$ to $E$ such that $BD = DE$. Join $EC$.
In $\Delta ADB$ and $\Delta CDE$:
$AD = CD$ (Since $D$ is the mid-point of $AC$)
$\angle ADB = \angle CDE$ (Vertically opposite angles)
$BD = DE$ (By construction)
Therefore,$\Delta ADB \cong \Delta CDE$ (By $SAS$ congruence criterion).
Thus,$AB = EC$ $(CPCT)$ and $\angle 1 = \angle 2$ $(CPCT)$.
Since $\angle 1$ and $\angle 2$ are alternate interior angles,$EC \parallel BA$.
Now,$EC \parallel BA$ and $BC$ is the transversal,so $\angle ABC + \angle BCE = 180^{\circ}$.
Since $\angle ABC = 90^{\circ}$,$\angle BCE = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
In $\Delta ABC$ and $\Delta ECB$:
$BC = CB$ (Common side)
$AB = EC$ (Proved above)
$\angle ABC = \angle ECB = 90^{\circ}$
Therefore,$\Delta ABC \cong \Delta ECB$ (By $SAS$ congruence criterion).
Thus,$AC = EB$ $(CPCT)$.
Since $BD = DE$,$BD = \frac{1}{2} BE$.
Substituting $BE = AC$,we get $BD = \frac{1}{2} AC$.