Theorem I, II and III — lesson. Mathematics CBSE, Class 9.

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Theorem I:

A diagonal of a parallelogram divides it into two congruent triangles.

Given:

$ABCD$ is a parallelogram, and

$AC$ is a diagonal.

Diagonal divides the quadrilateral into two triangles

$ABC$ and

$ADC$ .

To prove:

$\Delta ABC \cong \Delta CDA$

Proof: We know that "the opposite sides of a parallelogram are parallel".

$AB \ || \ DC$ and

$BC \ || \ AD$

Since

$AB \ || \ DC$ and

$AC$ is a transversal,

$\angle BAC = \angle DCA$ [alternate interior angles] - - - - - (I)

Also,

$BC \ || \ AD$ and

$AC$ is a transversal,

$\angle BCA = \angle DAC$ [alternate interior angles] - - - - - (II)

$\Delta ABC$ and

$\Delta CDA$ :

$\angle BAC = \angle DCA$ [from (I)]

$\angle BCA = \angle DAC$ [from (II)]

$AC = AC$ [Common side]

Thus,

$\Delta ABC \cong \Delta CDA$ [by

$ASA$ congruence rule].

Therefore, the diagonal of a parallelogram divides it into two congruent triangles.

Theorem II:

In a parallelogram, opposite sides are equal.

Given:

$ABCD$ is a parallelogram, and

$AC$ is a diagonal.

Diagonal divides the quadrilateral into two triangles

$ABC$ and

$ADC$ .

To prove:

$AB = CD$ and

$BC = AD$

Proof: We know that "the opposite sides of a parallelogram are parallel".

$AB \ || \ DC$ and

$BC \ || \ AD$

Since

$AB \ || \ DC$ and

$AC$ is a transversal,

$\angle BAC = \angle DCA$ [alternate interior angles] - - - - - (I)

Also,

$BC \ || \ AD$ and

$AC$ is a transversal,

$\angle BCA = \angle DAC$ [alternate interior angles] - - - - - (II)

$\Delta ABC$ and

$\Delta CDA$ :

$\angle BAC = \angle DCA$ [from (I)]

$\angle BCA = \angle DAC$ [from (II)]

$AC = AC$ [Common side]

Thus,

$\Delta ABC \cong \Delta CDA$ [by

$ASA$ congruence rule].

Corresponding parts of congruent triangles are equal.

$\Rightarrow AB = DC$ and

$BC = AD$ .

Therefore, the opposite sides of a parallelogram are equal.

Theorem III:

If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Given:

$ABCD$ is a quadrilateral, where

$AB = DC$ and

$BC = AD$ .

Construction: Join the diagonal

$AC$ .

To prove:

$ABCD$ is a parallelogram.

Proof: In

$\Delta ABC$ and

$\Delta CDA$ :

$AB = DC$ [Given]

$BC = AD$ [Given]

$AC = AC$ [Common side]

Thus,

$\Delta ABC \cong \Delta CDA$ [by

$SSS$ congruence rule].

$\implies \angle BAC = \angle DCA$ [by CPCT] - - - - (I)

$\implies \angle BCA = \angle DAC$ [by CPCT] - - - - (II)

These angles are alternate interior angles.

We know that "the alternate interior angles are equal only when the lines are parallel".

So,

$AB \ || \ DC$ and

$BC \ || \ AD$ .

Hence,

$ABCD$ is a parallelogram.

Important!

CPCT - Corresponding Parts of Congruence Triangles

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2. Theorem I, II and III

Theory: