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:

(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 $\Delta BMD$,

$AM=BM$ ($M$ is the mid-point of $AB$)

$\angle AMC=\angle BMD$ (Vertically opposite angles)

$CM=DM$ (Given)

$\therefore \Delta AMC\cong \Delta BMD$ (By SAS congruence rule)

$\therefore AC=BD$ (By CPCT)

And, $\angle ACM=\angle BDM$ (By CPCT)

(ii) $\angle ACM=\angle BDM$

However, $\angle ACM$ and $\angle BDM$ are alternate interior angles.

Since alternate angles are equal,

It can be said that $DB\parallel AC$

$\Rightarrow \angle DBC+\angle ACB={180}^{\circ }$ (Co-interior angles)

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

$\Rightarrow \angle DBC={90}^{\circ }$

(iii) In $\Delta DBC$ and $\Delta ACB$,

$DB=AC$ (Already proved)

$\angle DBC=\angle ACB$ (Each ${90}^{\circ }$)

$BC=CB$ (Common)

$\therefore \Delta DBC\cong \Delta ACB$ (SAS congruence rule)

(iv) $\Delta DBC\cong \Delta ACB$

$\therefore AB=DC$ (By CPCT)

$\Rightarrow AB=2CM$

$\therefore CM=\frac{1}{2}AB$