$ \mathrm{ABC} $ is a triangle right angled at $ \mathrm{C} $. A line through the mid-point $ \mathrm{M} $ of hypotenuse $ \mathrm{AB} $ and parallel to $ \mathrm{BC} $ intersects $ \mathrm{AC} $ at $ \mathrm{D} $. Show that
(i) $ \mathrm{D} $ is the mid-point of $ \mathrm{AC} $
(ii) $ \mathrm{MD} \perp \mathrm{AC} $
(iii) $ \mathrm{CM}=\mathrm{MA}=\frac{1}{2} \mathrm{AB} $

Given:

\( \mathrm{ABC} \) is a triangle right angled at \( \mathrm{C} \). A line through the mid-point \( \mathrm{M} \) of hypotenuse \( \mathrm{AB} \) and parallel to \( \mathrm{BC} \) intersects \( \mathrm{AC} \) at \( \mathrm{D} \).

To do:
We have to show that 

(i) \( \mathrm{D} \) is the mid-point of \( \mathrm{AC} \)
(ii) \( \mathrm{MD} \perp \mathrm{AC} \)
(iii) \( \mathrm{CM}=\mathrm{MA}=\frac{1}{2} \mathrm{AB} \)

 Solution:

\( \mathrm{ABC} \) is a triangle right angled at \( \mathrm{C} \).

This implies,

$\angle C=90^o$

$M$ is the mid-point of hypotenuse $AB$.

$DM \| BC$

(i) In $\triangle \mathrm{ABC}$,

$\mathrm{BC} \| \mathrm{MD}$ 

$\mathrm{M}$ is the mid-point of $\mathrm{AB}$.

Therefore, by converse of mid-point theorem, we get,

$D$ is the mid-point of $AC$.

(ii) $MD \| B C$ and $C D$ is the  transversal.

This implies,

$\angle A D M=\angle A C B=90^{\circ}$             (Corresponding angles are equal)

$\Rightarrow \mathrm{MD} \perp \mathrm{AC}$

(iii) In $\triangle A D M$ and $\triangle C D M$,

$\mathrm{DM}=\mathrm{DM}$          (Common side)

$AD=C D$        ($D$ is the mid-point of $AC$)

$\angle \mathrm{ADM}=\angle \mathrm{MDC} = 90^{\circ}$

Therefore, by SAS congruency, we get,

$\triangle \mathrm{ADM}=\triangle \mathrm{CDM}$

This implies,

$\mathrm{CM}=\mathrm{AM}$              (CPCT)......(i)

$M$ is the mid-point of $AB$.

$\mathrm{AM}=\mathrm{BM}=\frac{1}{2} \mathrm{AB}$.......(ii)

From (i) and (ii), we get,

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

Hence proved.

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Updated on: 10-Oct-2022

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