Question

In right angled 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 $B$. Show that:
(i) $△AMC\cong △BMD$
(ii) $\angle DBC$ is a right angle.
(iii) $△DBC\cong △ACB$
(iv) $CM=\frac{1}{2}AB$

Answer 1

(i) In $△AMC$ and $△BMD$, we have

$AM=BM$ $\mid$ Since $M$ is the mid- point of $AB$

$\angle AMC=\angle BMD$ $\mid$ Vertically opp. angles

and $CM=MD$ $\mid$ Given

$\therefore$ By SAS criterion of congruence, we have

$△AMC\cong △BMD$

(ii) Now, $△AMC\cong △BMD$

$\Rightarrow BD=CA$ and $\angle BDM=\angle ACM$ ... (1) $\mid$ Since corresponding parts of congruent triangles are equal

Thus, transversal $CD$ cuts $CA$ and $BD$ at $C$ and $D$ respectively such that the alternate angles $\angle BDM,\angle ACM$ are equal.

$\therefore ,BD\parallel CA.$

$\Rightarrow \angle CBD+\angle BCA={180}^{\circ }$ $\mid$ Since sum of consecutive interior angles are supplementary

$\Rightarrow \angle CBD+{90}^{\circ }={180}^{\circ }$ $\mid$ Since $\angle BCA={90}^{\circ }$

$\Rightarrow \angle CBD={180}^{\circ }-{90}^{\circ }$

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

(iii) Now, in $△DBC$ and $△ACB$, we have

$BD=CA$ $\mid$ From (1)

$\angle DBC=\angle ACB$ $\mid$ Since Each $={90}^o$

$BC=BC$ $\mid$ Common

$\therefore$ by SAS criterion of congruence, we have

$△DBC\cong △ACB$

(iv) $CD=AB$ [Since corresponding parts of congruent triangles are equal]

$M$ is the midpoint of $CD.$

$\Rightarrow CM=\frac{1}{2}CD$

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