Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half of the hypotenuse.

To do:

We have to prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half of the hypotenuse.

Solution:

Let in a right-angled triangle $ABC$,

$\angle B = 90^o$

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

Join $DB$.

Draw a circle with centre $D$ and $AC$ as diameter.

$\angle ABC = 90^o$

The circle drawn on $AC$ as diameter will pass through $B$

This implies,

$BD$ is the radius of the circle.

$AC$ is the diameter of the circle and $D$ is the mid-point of $AC$.

Therefore,

$AD = DC = BD$

$\Rightarrow BD= \frac{1}{2}AC$

Hence proved.

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Updated on: 10-Oct-2022

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