In the adjoining figure, ABC is a triangle right angled at C. A line segment through the mid-point M of AB and parallel to BC intersects AC at D. Show that CM = $\frac{1}{2}$ AB.

Open in App

Solution

In $△ A B C$ ,

$M$ is the mid-point of $A B$ and $M D ∥ B C$

$∴ D$ is the mid-point of $A C$

In $△ A M D$ and $△ C M D$

$A D = C D$ since $D$ is the mid-point of $A C$

$∠ A D M = ∠ C D M$ as $M D ⊥ A C$

$D M = D M$ (common)

$∴ △ A M D ≅ △ C M D$ by SAS congruence rule
$∴ A M = C M$ by CPCT ......... $(1)$

However, $A M = \frac{1}{2} A B$ since $M$ is the mid-point of $A B$ ......... $(2)$
From $(1)$ and $(2)$

$\Rightarrow C M = A M = \frac{1}{2} A B$
Hence proved.

Suggest Corrections

Similar questions

A B C

is a triangle right angled at

C

. A line through the mid-point

M

of hypotenuse

A B

and parallel to

B C

intersects AC at D. Show that

C M = M A = \frac{1}{2} A B

[4 MARKS]

$A B C$ is a triangle right angled at $C$ . A line through the mid-point $M$ of hypotenuse $A B$ and parallel to $B C$ intersects AC at D. Show that
$C M = M A = \frac{1}{2} A B$

Q. ABC is a triangle right angled at C. A line through the mid point M of hypotenuse AB and parallel to BC intersect AC at D .Show that :

C M = M A = \frac{1}{2} A B

Q. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallels to BC intersect AC at D. Show that CM=MA=

\frac{1}{2} A B

Question 7 (iii)
$A B C$ is a triangle right angled at $C$ . A line through the mid-point $M$ of hypotenuse $A B$ and parallel to $B C$ intersects AC at D. Show that
$C M = M A = \frac{1}{2} A B$ .

Join BYJU'S Learning Program

In the adjoining figure, ABC is a triangle right angled at C. A line segment through the mid-point M of AB and parallel to BC intersects AC at D. Show that CM = 12 AB.

In the adjoining figure, ABC is a triangle right angled at C. A line segment through the mid-point M of AB and parallel to BC intersects AC at D. Show that CM = $\frac{1}{2}$ AB.