Question

plz solve this fast​

Answer 1

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In Fig. 9.25, 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]

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Let us draw DN perpendicular to AC and BM perpendicular to AC.

${}{ \huge{ \blue{</strong><strong>(</strong><strong>i)</strong><strong>.}}}$

In ΔDON and ΔBOM,

• ∠DNO = ∠BMO ( 90°)

• ∠DON = ∠BOM (Vertically opposite angles)

• OD = OB (Given)

By AAS congruence rule,

ΔDON ΔBOM

DN = BM (BY CPCT) ..... (1)

We know that congruent triangles have equal areas.

Area (ΔDON) = Area (ΔBOM) ..... (2)

In ΔDNC and ΔBMA,

∠DNC = ∠BMA (Angles made by perpendicular)

CD = AB (Given)

DN = BM [Using equation (i)]

ΔDNC ΔBMA (RHS congruence rule)

Area (ΔDNC) = Area (ΔBMA) (iii)

Let us draw DN AC and BM AC

On adding equations (ii) and (iii)

Area (ΔDON) + Area (ΔDNC) = Area (ΔBOM) + Area (ΔBMA)

Therefore,

Area (ΔDOC) = Area (ΔAOB).

________________________

${}{ \huge{ \blue{(ii).}}}$

${}{ \tt{ \underline{ \red{To \: Prove - }}}}$

• Area (ΔDCB) = Area (ΔACB)

we have proved,

Area (ΔDOC) = Area (ΔAOB)

Area (ΔDOC) + Area (ΔOCB) = Area (ΔAOB) + Area (ΔOCB)

(Adding Area (ΔOCB) to both sides)

Area (ΔDCB) = Area (ΔACB).

__________________________

${}{ \huge{ \blue{(iii).}}}$

We have proved above,

Area (ΔDCB) = Area (ΔACB)

We know that,

[ If two triangles have the same base and equal areas, then these will lie between the same parallels ] .

So, DA || CB (iv)

In quadrilateral ABCD, one pair of opposite sides is equal (AB = CD) and the other pair of opposite sides is parallel (DA || CB)

Therefore, ABCD is a parallelogram.

__________________PROVED..

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