### Theory:

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 $$∠ODC = ∠OBA$$.

In $$\Delta AOB$$ & $$\Delta COD$$, 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.