Theorems IV and V — lesson. Mathematics CBSE, Class 9.

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Theorem IV

In a parallelogram, opposite angles are equal.

Given:

$ABCD$ is a parallelogram.

Construction: Join the diagonal

$AC$ .

To prove:

$\angle A = \angle C$ and

$\angle B = \angle D$

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

$AB \ || \ CD$ and

$BC \ || \ AD$

Since

$AB \ || \ CD$ and

$AC$ is a transversal.

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

Also,

$AD \ || \ BC$ and

$AC$ is a transversal.

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

Add equations (I) and (II).

$\angle BAC + \angle DAC = \angle DCA + \angle BCA$

$\angle BAD = \angle DCB$

That is,

$\angle A = \angle C$ - - - - - - (III)

Now, join the diagonal

$BD$ .

Since

$AB \ || \ CD$ and

$BD$ is a transversal.

$\angle ABD = \angle CDB$ [alternate interior angles] - - - - - (IV)

Also,

$AD \ || \ BC$ and

$BD$ is a transversal.

$\angle CBD = \angle ADB$ [alternate interior angles] - - - - - (V)

Add equations (IV) and (V).

$\angle ABD + \angle CBD = \angle CDB + \angle ADB$

$\angle ABC = \angle ADC$

That is,

$\angle B = \angle D$ - - - - - - (VI)

From (III) and (VI), we get:

The opposite angles of a parallelogram are equal.

Theorem V

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

Given:

$ABCD$ is a quadrilateral, where

$\angle A = \angle C$ and

$\angle B = \angle D$ .

To prove:

$ABCD$ is a parallelogram.

Proof: By angle sum property of a quadrilateral:

$\angle A + \angle B + \angle C + \angle D = 360^\circ$

$\angle A + \angle B + \angle A + \angle B = 360^\circ$ (Using given)

$2 \angle A + 2 \angle B = 360^\circ$

$2 ( \angle A + \angle B) = 360^\circ$

$\angle A + \angle B = 180^\circ$ - - - - (I)

Now, consider

$AB$ is the transversal for the lines

$AD$ and

$BC$ .

$\angle A$ and

$\angle B$ are interior angles on the same side of transversal line

$AB$ .

"Two lines cut by a transversal are parallel if and only if sum of interior angles on the same side of the transversal are

$180^\circ$ ".

Since

$\angle A + \angle B = 180^\circ$

$\Rightarrow AD \ || \ BC$ - - - - - (II)

Similarly,

$\angle A + \angle B + \angle C + \angle D = 360^\circ$

$\angle A + \angle D + \angle A + \angle D = 360^\circ$ (Using given)

$2 \angle A + 2 \angle D = 360^\circ$

$2 ( \angle A + \angle D) = 360^\circ$

$\angle A + \angle D = 180^\circ$ - - - - (III)

Similarly,

$AD$ is the transversal for lines

$AB$ and

$DC$ .

$\angle A$ and

$\angle D$ are interior angles on the same side of transversal line

$AD$ .

"Two lines cut by a transversal are parallel if and only if sum of interior angles on the same side of the transversal are

$180^\circ$ ".

Since

$\angle A + \angle D = 180^\circ$

$\Rightarrow AB \ || \ DC$ - - - - - (IV)

From equations (II) and (IV), we get:

$AD \ || \ BC$ and

$AB \ || \ DC$

Both pair of opposite sides are parallel.

Therefore,

$ABCD$ is a parallelogram.

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3. Theorems IV and V

Theory: