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Home / Geometry / Prove that opposite angles of a quadrilateral are equal to each other then it is a parallelogram

# Prove that opposite angles of a quadrilateral are equal to each other then it is a parallelogram

## Problem: Prove that opposite angles of a quadrilateral are equal to each other then it is a parallelogram.

General enunciation: we have to prove that, if opposite angles of a quadrilateral are equal to each other then it is a parallelogram.

Particular enunciation: Let ABCD is a quadrilateral. Here ∠B=∠D and ∠A=∠C. We have to prove that ABCD is a quadrilateral.

Proof: As ∠A=∠C and ∠B=∠D

Again ∠A+∠B+∠C+∠D=4 right angles.

Or,2∠A+2∠B=4 right angles.

Or, ∠A+∠B=2right angles.

∴AD ∥ BC. At same way, AB ∥ DC.

∴ABCD is a parallelogram.(proved)

### Problem: Prove that two diagonals of a rhombus bisect each other at right angle.

General enunciation: We have to prove that two diagonals of a rhombus bisect each other at right angle.

Particular enunciation: Let, ABCD is a rhombus. It`s diagonals AC and BD interject each other at “O”.

We have to prove that, ABCD is a rhombus that means,

(1)AO=CO and BO=DO.

(2) ∠AOB-∠BCO =∠COD=∠AOD-1 right angle.

Proof: Here, AB∥DC and AC is their interjector.

∠BAC=alternate ∠ABD. Now in ∆ABD and ∆COD,

∠OAB=∠OCD [alternate]

∠OBA=∠ODC [alternate]

AB=DC.

So, ∆AOB≌∆COD

∴AO=CO and BO=DO.

Now in ∆AOB and ∆AOD,

AB=AD, BO=DO and AO is common side .

∴∆AOB=∆AOD.

So, ∠AOB=∠AOD=1 right angle  [as they are adjacent angles]

Again, ∠COD=opposite ∠AOB=1 right angle

And ∠BOC=opposite ∠AOD=1 right angle.

Now, ∠COD=opposite ∠AOB=1 right angle.

And ∠BOC=opposite ∠AOD=1 right angle.

∴∠AOB=∠AOD=∠BOC=∠COD=1 right angle. (proved)

# Problem: prove that sum of four angles of a quadrilateral is four right angles.

General Enunciation: We have to prove that; sum of four angles of a quadrilateral is four right angles.

Particular Enunciation: Let, ABCD is a quadrilateral. We have to prove that, ∠ABC+∠BCD+∠CDA+∠DAB=four right angles.

Construction: Let us join A, C.

Proof: In ∆ABC< ∠ABC+∠BAC+∠ACB=2 right angles……..(1)

Again in ∆ADC; ∠ADC+∠ACD+∠CAD=2 right angles………(2)

Adding equation (1) and (2) we get,

∠ABC+∠BAC+∠ACB+∠ADC+∠ACD+∠CAD=(2+2)right angles.

Or, ∠ABC+(∠BAC+∠CAD)+( ∠ACB+∠ACD)+ ∠ADC=4 right angles

Or, ∠ABC+∠BAD+∠ADC+∠BCD=4 right angles.

That means, ∠ABC+∠BCD+∠CAD+∠DAB=4 right angles.(Proved)

Problem: Given that in the quadrilateral ABCD, AB=CD and ∠ABD=∠BDC, prove that ABCD is a parallelogram.

Particular enunciation: Given that, in the quadrilateral ABCD,

AB=CD and ∠ABD=∠BDC.

We have to prove that, ABCD is a parallelogram.

Proof: As, AB=CD            [Given]

And ∠ADB=∠DBC these two angles are alternate. So AD∥BC.

We know, if two sides of a quadrilateral are equal parallel then the quadrilateral will be a parallelogram.

Now, in quadrilateral ABCD,

AB=CD and AB∥CD. So, ABCD is a parallelogram (Proved)

Problem: Given that in the parallelogram ABCD, BM is the bisector of ∠ABC and DN is the bisector of ∠ADC. Prove that BMDN is a parallelogram.

Particular enunciation: Given that, in the parallelogram BM is the bisector of ∠ABC and DN is the bisector of ∠ADC. Now join B ,N and D ,M.

We have to prove that, BMDN is a parallelogram.

Proof: Between, ∆AND and ∆BMC –

∠AND =∠MBC    [being half of the opposite angles of parallelogram]

∠DAN=∠BCM           [Alternative angle]

AD = BC [opposite sides of parallelogram are equal]

∴∆AND ≌∆BMC

∴DN=BM

Again the internal bisectors of opposite angles of parallelogram are parallel each other.

∴DN=BM.

So, DN and BM are equal parallel.

∴BMDN is a parallelogram.   (Proved)

Problem: Given that in the parallelogram ABCD, AP=CR and DS=QB. Prove that PQRS is a parallelogram.

Particular enunciation: Given that in the parallelogram ABCD,

AP=CR and DS=QB. We have to prove that, PQRS is a parallelogram.

Construction: Let us join P, R.

Proof: AP=CR             [given]

∴DP = BR

[Because of opposite sides of parallelogram AD=BC]

Again, DS=QB

∴CS=AQ  [opposite sides of parallelogram AB=DC]

Now in ∆APQ and ∆RSC

AP=CR

AQ=CS

Interior ∠QAP=interior ∠SCR

∴∆APQ and ∆RSC are congruent.

∴PQ=RS

Again in∆ DPS and∆ BQR-

DP=BR

DS=BQ

And interior ∠PDS=interior ∠RBQ

∴  APQ  ≌RSC

∴ PS= QR

∴ PQRS is a parallelogram (Proved)

## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

If two triangles have the three sides of the one equal to the three sides ...