# $\mathrm{ABC}$ 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) $\mathrm{CM}=\mathrm{MA}=\frac{1}{2} \mathrm{AB}$

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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.

Updated on 10-Oct-2022 13:41:07