Question

"Question 6 In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that: (i) ar (DOC) = ar (AOB) (ii) ar (DCB) = ar (ACB) (iii) DA || CB or ABCD is a parallelogram. [Hint: From D and B, draw perpendiculars to AC.]

Class 9 - Math - Areas of Parallelograms and Triangles Page 163"

Answer 1

Parallelogram :

A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram.

 

Two Triangles having the same base and equal areas lie between the same parallels.

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Given, ABCD is a quadrilateral in which AB = CD its diagonals AC and BD intersect at O such  that OB=OD.

To show:

(i) ar (DOC) = ar (AOB)

(ii) ar (DCB) = ar (ACB)

(iii) DA || CB or ABCD is a parallelogram.

Construction,
DE ⊥ AC and BF ⊥ AC are drawn.

Proof:
(i) In ΔDOE and ΔBOF,

∠DEO = ∠BFO (each 90°)

∠DOE = ∠BOF (Vertically opposite angles)

OD = OB (Given)

Therefore, ΔDOE ≅ ΔBOF

(by AAS congruence rule)

Thus, DE = BF (By CPCT) — (i)

also, ar(ΔDOE) = ar(ΔBOF) ........(ii)

(Two Congruent triangles have equal areas)

Now,
In ΔDEC and ΔBFA,

∠DEC = ∠BFA (each 90°)

CD = AB (Given)

DE = BF (From i)

Therefore,ΔDEC ≅ ΔBFA

byy RHS congruence rule)

Thus, ar(ΔDEC) = ar(ΔBFA) ........(iii)

(Two Congruent triangles have equal areas)

Adding (ii) and (iii),

ar(ΔDOE) + ar(ΔDEC) = ar(ΔBOF) + ar(ΔBFA)

 ar (DOC) = ar (AOB)

 

(ii)  ar(ΔDOC) = ar(ΔAOB)

⇒ ar(ΔDOC) + ar(ΔOCB) = ar(ΔAOB) + ar(ΔOCB)   

(Adding ar(ΔOCB) on both sides)

 ar(ΔDCB) = ar(ΔACB)  

 

(iii) From part (ii) (ΔDCB) & (ΔACB) have all areas and have the same base BC. So, (ΔDCB) & (ΔACB) must lie between the same parallels.

 DA || BC — (iv)

∠FBO= ∠EDO.....(v)

(ΔDOE ≅ ΔBOF )

∠FBA=∠EDC.....(vi)

(ΔDEC ≅ ΔBFA )

 On adding eq v & vi

∠ABD=∠CDB

Therfore, DC||AB......(vii)

From eq iv & vii, We get DA||CB & DC||AB

Hence, ABCD is parallelogram  .

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Hope this will help you...

 

Answer 2

The diagram is in the attached image.

 Draw a line EO such that

EO || AB

 EADE = BODO  ….(i) [ Proportionality Theorem]

Also,  BOAO =  DOCO Given

BODO  = AOCO ...(ii) 

From equation (i) and (ii), we get

EADE = AOCO

By using converse of Basic Proportionality Theorem, EO || DC also EO || AB

 AB || DC.

Hence, quadrilateral ABCD may be a trapezium with AB || CD or may be a parallelogram.