CAT Theorems of a Parallelogram - Practice Questions & MCQ

Before moving to the theorems and proofs, let's do an activity. Cut out a parallelogram from a sheet of paper and cut it along a diagonal. You obtain two triangles.  What do you observe? Place one triangle over the other. Turn one around, if necessary. Note down your observations and after studying the theorems

Theorem 1: A diagonal of a parallelogram divides it into two congruent triangles.

Let ABCD be a parallelogram and AC be a diagonal. Observe that the diagonal AC divides parallelogram ABCD into two triangles, namely, ∆ ABC and ∆ ACD. And, we need to prove that these triangles are congruent.

In ∆ ABC and ∆ ACD, note that BC || AD and AC is a transversal.

So,         ∠ BCA = ∠ DAC                 (Pair of alternate angles)

Also,      AB || DC and AC is a transversal.

So,         ∠ BAC = ∠ DCA                 (Pair of alternate angles)

and         AC = CA                             (Common)

So,         ∆ ABC ≅ ∆ ACD                 (ASA rule)

or, diagonal AC divides parallelogram ABCD into two congruent triangles ABC and ACD.

Theorem 2: In a parallelogram, opposite sides are equal.

As we have already proved that a diagonal divides the parallelogram into two congruent triangles.

Now, in ∆ ABC and ∆ ACD, the corresponding sides are equal

So, AB = DC and AD = BC

The converse of above theorem is-

Theorem 3: If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Let sides AB and CD of the quadrilateral ABCD be equal and also AD = BC. Draw diagonal AC.

In ∆ ABC and ∆ ACD, we have

                            AB = CD                (given)

                            BC = AD                (given)

                            AC = CA                (common)

Therefore,     ∆ ABC ≅ ∆ ACD

Therefore,     ∠ BAC = ∠ DCA

and               ∠ BCA = ∠ DAC

But, these are alternate interior angles.

Therefore,    AB || CD  and similarly,  AD || BC.

Hence, ABCD is a parallelogram.

Theorem 4 : In a parallelogram, opposite angles are equal.

Let ABCD be the parallelogram

So,                ∆ ABC ≅ ∆ CDA             (proved, ASA rule)

                          ∠ B = ∠ D                  (Corresponding parts of Congruent triangles)

Also,             ∠ BAC = ∠ DCA   and     ∠ BCA = ∠ DAC           

Therefore,    ∠ BAC + ∠ DAC = ∠ DCA + ∠ BCA

i.e.                     ∠ A = ∠ C

Hence, ∠ A = ∠ C  and ∠ B = ∠ D.

The converse of the above theorem-

Theorem 5 : If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.

Let ABCD is a quadrilateral in which ∠ A = ∠ C  and ∠ B = ∠ D.

We need to prove ABCD is a parallelogram.

We have,

                     ∠ A = ∠ C  and ∠ B = ∠ D          (Given)

⇒                 ∠ A + ∠ B = ∠ C + ∠ D

⇒                 ∠ A + ∠ B = ∠ C + ∠ D = 180°               (∵ ∠ A + ∠ B + ∠ C + ∠ D = 360°)

Now, the line segments BC and AD are cut by the transversal AB such that ∠ A + ∠ B = 180° 

Therefore,      AC || BD                                            ( ∵ ∠ A and ∠ B are co interior angle)

Again,           ∠ A = ∠ C  and ∠ D = ∠ B 

⇒                  ∠ A + ∠ D = ∠ C + ∠ B

⇒                 ∠ A + ∠ D = ∠ C + ∠ B = 180°           (∵ ∠ A + ∠ B + ∠ C + ∠ D = 360°)

Now, the line segments BC and AD are cut by the transversal AB such that ∠ A + ∠ D = 180° 

Therefore,      AB || DC                                          ( ∵ ∠ A and ∠ D are co interior angle)

Thus,  AB || DC and AD || BC

Hence, ABCD is a parallelogram.

Theorem 6 : The diagonals of a parallelogram bisect each other.

Now, if in a quadrilateral the diagonals bisect each other, will it be a parallelogram?

Indeed this is true and this result is the converse of the above theorem.

Theorem 7 : If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Let us prove this

We have given a quadrilateral  ABCD whose diagonals AC and BD intersect at a point O such that OA = OC and OB = OD.

We have to prove that ABCD is a parallelogram.

In ∆ AOB and ∆ COD, we have

                            OA = OC and OB = OD                 (given)

and,               ∠ AOB = ∠ COD                                 (vertically opposite angle)

Therefore,     ∆ AOB ≅ ∆ COD                                 (SAS criteria)

Therefore,     ∠ BAO = ∠ DCO

But, these are alternate interior angles.

From this, we get AB || CD

Again, in ∆ AOD and ∆ COB, we have

                            OA = OC and OB = OD                 (given)

and,               ∠ AOD = ∠ COB                                 (vertically opposite angle)

Therefore,     ∆ AOD ≅ ∆ COB                                 (SAS criteria)

Therefore,     ∠ ADO = ∠ CBO

But, these are alternate interior angles.

From this, we get AD || BC

Thus, AB || CD and AD || BC

Hence, ABCD is a parallelogram.

Till now we have studied many properties and theorems of a parallelogram and we have also verified that if in a quadrilateral any one of those properties is satisfied, then it becomes a parallelogram.

We now study yet another condition which is the least required condition for a quadrilateral to be a parallelogram.

Theorem 8 : A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

Let's prove this

We have given a quadrilateral in which AB = DC and AB || DC.

We need to prove ABCD is a parallelogram.

Draw a diagonal AC

n ∆ ABC and ∆ CDA, we have

                          AB = CD                   (given)

                          AC = CA                   (common)

and              ∠ BAC = ∠ DCA             (alternate interior angle as AB || DC and CA cuts them)

Therefore,   ∆ ABC ≅ ∆ CDA             (SAS criteria)

Therefore,   ∠ BCA = ∠ DAC

But, these are alternate interior angles.

Therefore,         AD || BC

Thus, AB || DC and AD || BC

Hencem ABCD is a parallelogram.