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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.

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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.

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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.

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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.

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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.

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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.

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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)

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