Question

In right triangle $ABC$, right angled at $C,M$ is the mid-point of hypotenuse $AB$. $C$ is joined to $M$ and produced to a point $D$ such that $DM=CM$. Point $D$ is joined to point $B$ (see figure).
Show that:
(i) $\Delta AMC\cong \Delta BMD$
(ii) $\angle DBC$ is a right angle.
(iii) $\Delta DBC\cong \Delta ACB$
(iv) $CM=\frac{1}{2}AB$

Answer 1

(i) In $\Delta AMC$ and $\Delta BMD$
$DM=CM$ (given)
$\angle 3=\angle 4$ (vertically opposite angles)
$AM=MB$ (as $M$ is the mid-point of $AB$)
$\Delta AMC\cong \Delta BMD$ (by SAS congruency rule)
(ii) $\because \Delta AMC\cong \Delta BMD$
$\angle 1=\angle 2$ (by c.p.c.t)
But they are alternate angles,
so $AC||BD$
$\angle ACB+\angle DBC={180}^0$
(sum of the interior angles on the same side of a transversal is equal to ${180}^0$)
But $\angle ACB={90}^0$ (given)
$\angle DBC={180}^0={90}^0$
(iii) In $\Delta DBC$ and $\Delta ACB$
$\angle DBC=\angle ACB={90}^0$
$BC=BC$ (common side)
$BD=CA$ $(\because \Delta AMC\cong \Delta BMD)$
$\therefore \Delta DBC\cong \Delta ACB$ (by SAS congruency rule)
(iv) $\because \Delta DBC\cong \Delta ACB$
$DC=AB$
But $M$ mid-point of $DC$
$2CM=CD$
$2CM=AB[\because DC=AB]$
$CM=\frac{1}{2}AB$.
Hence proved.