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

Given:

In right triangle ABC, right angledat C. M is the mid-point of hypotenuse AB. C is joined to \( \mathrm{M} \) and produced to a point \( \mathrm{D} \) such that \( \mathrm{DM}=\mathrm{CM} \). Point \( \mathrm{D} \) is joined to point \( \mathrm{B} \).

To do:

We have to show that

(i) \( \triangle A M C \cong \triangle B M D \)
(ii) \( \angle \mathrm{DBC} \) is a right angle
(iii) \( \triangle \mathrm{DBC} \cong \triangle \mathrm{ACB} \)
(iv) $CM=\frac{1}{2}AB$

Solution:

$\angle ACB=90^o$

$AM=BM$

$DM=CM$

(i) In $\triangle AMC$ and $\triangle DMB$,

$AM=BM$

$CM=DM$

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

Therefore,

$\triangle AMC \cong\ \triangle DMB$    (By SAS congruence rule)

(ii) $\triangle AMC \cong\ \triangle DMB$

This implies,

$\angle ACM=\angle BDM$   (CPCT)

$\angle ACM$ and $\angle BDM$ are alternate angles to lines AC and BD.

This implies,

$AC \parallel\ BD$

$\angle ACB+\angle DBC=180^o$    (Angles on the same side of the transversal are supplementary)

$90^o+\angle DBC=180^o$

$\angle DBC=180^o-90^o$

$\angle DBC=90^o$

Therefore, $\angle DBC$ is a right angle.

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

$AC=DB$     ($\triangle AMC \cong\ \triangle DMB$, CPCT)

$BC=BC$     (Common side)

$\angle ACB=\angle DBC=90^o$

Therefore,

$\triangle ACB \cong\ \triangle DBC$    (By SAS congruence rule)

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

This implies,

$AB=DC$    (CPCT)

$\frac{1}{2}AB=\frac{1}{2}DC$....(i)

$DM=CM=\frac{1}{2}DC$....(ii)

From (i) and (ii),

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

Hence proved.