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 the given figure). Show that: $\left(i\right)△AMC\cong △BMD$ $\left(ii\right)CM=\frac{1}{2}AB$

Answer 1

Proof (i) In $\Delta AMC$ and $\Delta BMD$

M is the midpoint $\Rightarrow AM=BM$

$\Rightarrow \angle CMA=\angle DMB$ (vertically opposite angles)

$CM=DM$ (given)

Hence $\Delta AMC\cong \Delta BMD$ by $SAS$ congruence rule

(ii) $DC=AB$ Since $\Delta DBC\cong \Delta ACB$

$\Rightarrow DM+CM=AB$

$\Rightarrow CM+CM=AB$ Since $CD=CM+DM$

$\Rightarrow 2CM=AB$ and $CM=DM$

Hence $CM=1/2AB$