The Mid-point Theorem — lesson. Mathematics CBSE, Class 9.

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Theorem IX

The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Given:

$ABC$ is a triangle, where

$E$ is the mid-point of

$AB$ and

$F$ is the mid-point of

$AC$ .

To prove:

$EF \ || \ BC$

Construction: Draw a line segment through

$C$ parallel to

$AB$ and extend

$EF$ to meet this line at

$D$ .

Proof: Since

$E$ is the mid-point of

$AB$ ,

$AE = EB$ - - - - (I)

Since

$F$ is the mid-point of

$AC$ ,

$AF = FC$ - - - - (II)

By construction

$AB \ || \ CD$ .

$\Rightarrow AE \ || \ CD$ and

$ED$ is a transversal.

$\angle AEF = \angle CDF$ [alternate interior angles] - - - - - (III)

$\Delta AEF$ and

$\Delta CDF$ :

$\angle AFE = \angle CFD$ [vertically opposite angles]

$\angle AEF = \angle CDF$ [(from (III)]

$AF = FC$ [from (II)]

Therefore,

$\Delta AEF \cong \Delta CDF$ [by

$AAS$ congruence rule].

$\Rightarrow AE = CD$ [by CPCT] - - - - - (IV)

From equations (I) and (IV), we can say that:

$BE = CD$

In quadrilateral

$EBCD$ :

$BE \ || \ CD$ and

$BE = CD$

One pair of opposite side is equal and parallel.

Thus,

$EBCD$ is a parallelogram.

So,

$ED \ || \ BC$

$\Rightarrow EF \ || \ BC$

Hence, proved.

Note: We proved that

$\Delta AEF \cong \Delta CDF$ .

$EF = FD$ [by CPCT]

$EF = \frac{1}{2} ED$

$EF = \frac{1}{2} BC$ [Since

$EBCD$ is a parallelogram,

$ED = BC$ ].

Thus, we can conclude that the parallel mid-line segment is half of the third side of the triangle.

Important!
The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.

Theorem X

The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

Given:

$ABC$ is a triangle, where

$E$ is the mid-point of

$AB$ . Also,

$EF \ || \ BC$ .

Construction: Draw a line segment through

$C$ parallel to

$AB$ and extend

$EF$ to meet this line at

$D$ .

To prove:

$F$ is the mid-point of

$AC$ . That is,

$AF = FC$

Proof: Since

$E$ is the mid-point of

$AB$ ,

$AE = EB$ - - - - (I)

In quadrilateral

$EBCD$ :

$EF \ || \ BC$ [since

$ED \ || \ BC$ ]

$EB \ || \ DC$ [by construction]

Both pairs of opposite sides are parallel.

Thus,

$EBCD$ is a parallelogram.

We know that "opposite sides of a parallelogram are equal".

$EB = DC$ - - - - (II)

On comparing equation (I) and (II), we get,

$AE = DC$ - - - - (III).

Since

$AE \ || \ DC$ with transversal

$ED$ by construction.

$\angle AEF = \angle CDF$ [alternate interior angles] - - - - (IV)

$\Delta AEF$ and

$\Delta CDF$ :

$\angle AEF = \angle CDF$ [from (IV)]

$\angle AFE = \angle CFD$ [vertically opposite angles]

$AE = DC$ [from (III)]

Therefore,

$\Delta AEF \cong \Delta CDF$ [by

$AAS$ congruence criterion].

$\Rightarrow AF = CF$ [by CPCT]

Thus,

$F$ is the mid-point of

$AC$ .

Hence, proved.

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6. The Mid-point Theorem

Theory: