$A B C$ is a triangle right angled at $C$ . A line through the midpoint $M$ of hypotenuse $A B$ and Parallel to $B C$ intersects $A C$ at $D$ . Show that:
$M D ⊥ A C$

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Solution

As,

$M D ∥ B C$ and

$A C$ is traversal

$∴ ∠ M D C + ∠ B C D = 180 °$ (Interior angles on the same side of transversal are supplementary)

$∠ M D C + 90 ° = 180 °$

$∠ M D C = 90 °$

$∴ M D ⊥ A C$

Hence proved

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triangle ABC is right angled at C. A line through the midpoint M of hypotenuse AB and parallel to BC intersects Ac at D. Show that MD perpendicular to AC

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MD $⊥$ AC .

Q. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB an parallel to BC intersects AC at D. Show that
MD

⊥

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is a triangle right angled at

C

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M

of hypotenuse

A B

and Parallel to

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D

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D

is the midpoint of

A C

Q. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB an parallel to BC intersects AC at D. Show that
MD

⊥

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ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and Parallel to BC intersects AC at D. Show that:MD⊥AC

As,MD∥BC and AC is traversal∴∠MDC+∠BCD=180° (Interior angles on the same side of transversal are supplementary)∠MDC+90°=180°∠MDC=90°∴MD⊥ACHence proved

$A B C$ is a triangle right angled at $C$ . A line through the midpoint $M$ of hypotenuse $A B$ and Parallel to $B C$ intersects $A C$ at $D$ . Show that:
$M D ⊥ A C$

As,

$M D ∥ B C$ and

$A C$ is traversal

$∴ ∠ M D C + ∠ B C D = 180 °$ (Interior angles on the same side of transversal are supplementary)

$∠ M D C + 90 ° = 180 °$

$∠ M D C = 90 °$

$∴ M D ⊥ A C$

Hence proved