Theorem : In a parallelogram, opposite angles are equal.

 Given: A parallelogram ABCD,  opposite sides of parallelogram are 
                        side AB and side DC 
                        side AD and side BC.

To prove: Opposite angles of parallelogram are equal.

      that is ∠A = ∠C    and    ∠D = ∠B

Construction: Join A to C that is AC, is a diagonal, the diagonal AC divides parallelogram ABCD into two triangles △ ABC and △ CDA.

Proof: In △ ABC and △ CDA

BC ∥ AD and AC is a transversal.

So, ∠BCA = ∠DAC….(1) (Alternate angles of parallel sides)

Similarly, AB ∥ DC and AC is a transversal.

So, ∠DCA = ∠BAC ….(2) (Alternate angles of parallel sides)

                          Add (1) and (2)

              ∠BCA + ∠DCA = ∠DAC + ∠BAC

                           ∠BCD = ∠BAD

similarly               ∠ADC = ∠ABC

Hence proved.

Second method 

Given: A parallelogram ABCD,  opposite sides of parallelogram are 
                        side AB and side DC 
                        side AD and side BC.

To prove: Opposite angles of parallelogram are equal.

          that is ∠A = ∠C    and    ∠D = ∠B 

Proof:  BC ∥ AD and CD is a transversal.

So, ∠C + ∠D = 1800….(1) (Consecutive interior angles)

Similarly, AB ∥ DC and AD is a transversal.

So, ∠A + ∠D = 1800….(2) (Consecutive interior angles)

compare (1) and (2)

∠C + ∠D = ∠A + ∠D

∠A = ∠C

similarly, ∠D = ∠B


Hence proved.

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