Parallelogram theorem-III, IV, V — lesson. Mathematics State Board, Class 9.

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Theorem III: In parallelogram, opposite angles are equal.

Given: A parallelogram

$ABCD$ with

$AC$ and

$BD$ as its diagonals.

To prove:

$∠A = ∠C$ &

$∠B = ∠D$ .

Proof: We know that 'Opposite sides of a parallelogram is parallel and equal'. So,

$AB||DC$ and

$AD||BC$ .

Since

$AB||DC$ with its transversal line

$AC$ .

$∠ACD = ∠CAB$ .

Since

$AD||BC$ with its transversal line

$AC$ .

$∠DAC = ∠BCA$ .

Thus, $∠A = ∠DAC+∠CAB = ∠BCA+∠ACD = ∠C$.

Since

$AB||DC$ with its transversal line

$BD$ .

$∠CDB = ∠DBA$ .

Since

$AD||BC$ with its transversal line

$BD$ .

$∠ADB = ∠DBC$ .

Thus,

$∠B = DBA+∠DBC = ∠CDB+∠ADB = ∠D$ .

Hence, it is proved.

Theorem IV: The diagonal of a parallelogram bisect each other.

Given: A parallelogram

$ABCD$ with

$AC$ and

$BD$ and

$O$ is the point of intesection of

$AC$ and

$BD$ .

To prove:

$OA = OC$ and

$OB = OD$ .

Proof: Since 'Opposite sides of a parallelogram is parallel'. So,

$AB||DC$

$AD||BC$ .

Since

$AB||DC$ &

$AC$ is the transversal.

$∠ACD = ∠OCD = ∠CAB = ∠OAB$ (Alternate Interior angle ...1)

That is

$∠OCD = ∠OAB$ .

Since

$AB||DC$ &

$BD$ is the transversal.

$∠BDC = ∠ODC = ∠DBA = ∠OBA$ (Alternate Interior angle ...1)

That is

$∠ODC = ∠OBA$ .

$\Delta AOB$ &

$\Delta COD$ , two corresponding angles are equal (

$∠OCD = ∠OAB$ and

$∠ODC = ∠OBA$ ) and a side is equal in measure.

Thus, by the

$ASA$ criterion, the two triangles

$\Delta AOB$ &

$\Delta COD$ are congruent, which means that the corresponding sides must be equal.

Therefore,

$△AOB ≅ △ COD$ (ASA congruency)

So,

$OA = OC$ and

$OB = OD$ (Corresponding parts of Congruent triangles)

Hence, it is proved.

Theorem V: Parallelograms on the same base and between the same parallels are equal in area.

Given: Two parallelogram

$ABCD$ &

$ABEF$ that has same base

$AB$ & lies between same parallel

$CF$ &

$AB$ .

To prove: Area of parallelogram

$ABCD$

$=$ Area of parallelogram

$ABEF$

Proof: It can be seen that the parallelogram

$ABCD$ and

$ABEF$ are on the common base

$AB$ .

Here we have a pair of triangle

$ADF$ and

$BCE$ .

$\angle ADF = \angle BCD$ and

$\angle AFD \angle BEC$ .

We know that

$AD = BC$ and

$AF =BE$ (Opposite sides are equal in measure).

Thus, two of the sides and two of the angles are equal.

By AAS congruence, the triangles

$ADF$ and

$BCE$ are congruent triangles.

We know that the congruent triangles have the same area.

Now, area of parllelogram

$ABCD$

$=$ Area of

$\Delta BCE$

$+$ Area of quadrilateral

$ABED$ .

$=$

$\Delta BCE$

$+$ Area of parallelogram

$ABED$

$=$ Area of parallelogram

$ABEF$

Area of parllelogram

$ABCD$

$=$ Area of parallelogram

$EFCD$

Hence, it is proved.

In this process, we have also proved other interesting statements. These are called Corollaries, which do not need separate detailed proofs.

Important!

Corollary 1: Triangles on the same base and between the same parallels are equal in area.

Corollary 2: A rectangle and a parallelogram on the same base and between the same parallels are equal in area.

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4. Parallelogram theorem-III, IV, V

Theory: