Theorem – In a parallelogram, opposite angles are equal
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.