### Exercise: Prove that, if the opposite sides of a quadrilateral are equal and parallel, it is a parallelogram.

General enunciation: We have to prove that if the opposite sides of a quadrilateral are equal and parallel, it is a parallelogram.

Particular enunciation: Let the opposite sides of the quadrilateral ABCD are equal and parallel .i.e. BC=AD,BC∥AD and AB=DC,AB∥DC

We have to prove that ABCD is a parallelogram.

Construction: We join A,C and B,D.

Proof: In ∆ABC and ∆ADC,

AB=DC, BC=AD [supposition]

And AC = AC [common side]

∴∆ABC≌∆ADC

∴∠B=∠D [SSS theorem] ————————- (i)

Again, in ∆ABD and ∆BCD,

AB=DC, BC=AD [supposition]

And BD=BD [common side]

∴∆ABD≌∆BCD

∴∠A=∠C [SSS theorem] ——————————-(ii)

Now, in the quadrilateral ABCD,

BC=AD,AB=DC [from (i)and (ii)]

∴ ∠A =C, ∠B=∠D [∵the opposite sides and angles of a parallelogram are equal]

∴ABCD is a parallelogram [Proved]

**Exercise: Prove that, the bisectors of any two opposite angles of a parallelogram are parallel to each other.**

General enunciation: To prove that the bisectors of any two opposite angles of a parallelogram are parallel to each other.

Particular enunciation: Let ABCD is a parallelogram.

It’s the bisectors of any two opposite angles ∠ABC and ∠ADC are BF and DF respectively.

We have to prove that DE∥BF.

Proof: ABCD is a parallelogram.

So ∠ABC=∠ADC [∵opposite angles of a parallelogram are equal]

Since BF and DE are their bisectors respectively.

∴∠EBF=∠EDF ————————————— (i)

Again, AB∥CD and BF is their transversal

∴∠EBF=∠BFC ————————- (ii) [alternative angles are equal]

Now, ∠EDF=∠EBF=∠BFC

∴∠EDF=∠BFC [from (i) and (ii)]

But these are corresponding angles where DF is the transversal.

Therefore DE∥BF (Proved)

**Exercise: Prove that, the bisectors of any two adjacent angles of a parallelogram are perpendicular to each other.**

General enunciation: To prove that the bisectors of any two adjacent angles of a parallelogram are perpendicular to each other.

Particular enunciation: Let ABCD is a parallelogram. It’s the bisectors of two adjacent angles ∠ABC and ∠BCD are BO and CO respectively.

We have to prove that BO⊥CO.

Proof: ABCD is a parallelogram so AB∥CD and BC is their transversal

When a transversal cuts two parallel straight lines then the sum of the pair of interior angle on the same side of the transversal is 2 right angles.

∴∠ABC+∠BCD=180^{0 }

Or, ½ ∠ABC + ½ ∠BCD=90^{0}

Or, ∠OBC+∠OCB = 90^{0} —————————- (1)

In ∆BOC, we know that sum of three angles of a triangle is equal to 2 right angles

∴ ∠OBC+∠OCB+∠BOC=180^{0 }

Or, 90^{0}+∠BOC=180^{0} [from (1)]

Or, ∠BOC=180^{0}-90^{0}

Or, ∠BOC=90^{0}

Therefore BO⊥CO (Proved)