Question

mathmatic
marks -10
Diagonal Ac and BD of quadrilateral is
Intreset at o such that og op
i) AH.DOC = Ar AOB
in Ar.DOC AM.AOB​

Answer 1

Answer:

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Answer 2

Answer:

i)Let us draw DPperpendicular to AC and BQ perpendicular AC

In △DOP and △BOQ

∠DPO=∠BQO=90

∘

∠DOP=∠BOQ (vertically opposite angles)

OD=OB(given)

△DOP≅△BOQ(AAS congruence rule)

∴DP=BQ by CPCT .....(1)

and area(DOP)=area(BOQ) ......(2) since area of congruent triangles are equal.

In △CDP and △ABQ,

∠CPD=∠AQB=90

∘

CD=AB(given)

DP=BQ from (1)

∴△CDP≅△ABQ by RHS congruence rule.

⇒ area△CPD=area△AQB ........(3)since area of congruent triangles is equal.

Adding (2) and (3)

Area△(DOP)+Area△(CDP)=area△(BOQ)+area△(ABQ)

⇒ Area△(DOC)=Area△(AOB)

Hence proved.

In part (i) we proved that

Area△(DOC)=Area△(AOB)

Adding Area△(OCB) both sides,

Area△(DOC)+Area△(OCB)=Area△(AOB)+Area△(OCB)

⇒ Area△(DCB)=area△(ACB)

In part (ii) we proved

Area△(DCB)=area△(ACB)

We know that two triangles having same base and equal areas, lie between same parallels.

Here △DBC and △ABC are on the same base BC and are equal in area,

So, these triangles lie between the same parallels DA and CB

∴DA∥CB

Now, in △DOA and △BOC

∠DOA=∠BOC(vertically opposite angles)

∠DAO=∠BOC since DA∥CB,AC transversal, alternate angles are equal.

OD=OB(given)

∴△DOA≅△BOC (AAS congruency)

DA=BC by CPCT.

So, in ABCD, since DA∥CB and DA=CB

One pair of opposite sides of quadrilateral ABCD is equal and parallel.

∴ABCD is a parallelogram.

Explanation:

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