Question

Hiya!!!

IN FIGURE DIAGONALS ,AC AND BD OF QUADRILATERAL ABCD INTERSECT AT O SUCH THAT,OB=OD.

IF AB=CD,SHOW THAT
1. AREA OF (DOC)=AREA OF (AOB)
2.AREA OF (DCB)=AREA OF (ACB)
3.ABCD IS A PARALLELOGRAM.

hint:construction:draw DL PERPENDICULAR AC AND BM PERPENDICULAR AC.

SPAMSSSS WILL BE REPORTED.

ONLY MYSTERY ANSWERS BE MARKED BRAINLIEST ♥♥♥

Answer 1

$\large\boxed{\large\boxed{\mathbb{ELLO \ THERE!}}}}}$

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$\underline{\mathsf{ANSWER}}$

Given

ABCD is a quadrilateral

OB = OD

AB = CD

To Prove

i) ar[DOC] = ar[AOB]

ii) ar[DCB] = ar[ACB]

iii) ABCD is a parallelogram

Construction

Draw Dl ⊥ AC and BM ⊥ AC

Proof

In ΔBOM and ΔDOL,

OB = OD [Given]

∠BMO = ∠DLO = 90°

∠BOM = ∠DOL [V.O.A]

∴ ΔBOM ≅ ΔDOL by AAS congruency.

ar[BOM] = ar[DOL] → 1

[Congruent figures have equal areas]

$\Longrightarrow$ BM = DL [CPCT]  

In ΔBAM and ΔDCL,

BM = DL [proved]

AB = CD [given]

∠BMA = ∠DLC = 90°

∴ ΔBAM ≅ ΔDCL by RHS congruency.

∠BAM = ∠DCL [CPCT]

$\Longrightarrow$  AB║DC [Alternate interior anls are equal.]

ar[BAM] = ar[DCL] → 2

i) Adding 1 and 2 we get,

ar[BOM] + ar[BAM] = ar[DOL] + ar[DCL]

ar[AOB] = ar[DOC]

ii) ar[AOB] = ar[DOC]

Adding BOC on both sides.

ar[AOB] + ar[BOC] = ar[DOC] + ar[BOC]

ar[ACB] = ar[DCB]

iii) AB = CD [given]

AB║CD [proved]

⇒ ABCD is a parallelogram.

[One pair of opposite sides are equal and parallel]

Hence everything is Proved!

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Hope it Helps!

@Tomboyish44

Answer 2

Step-by-step explanation:

Given ABCD is a quadrilateral with AB = CD.Diagonals AC and BD intersect at O such that OB = OD.

Construction: Draw DN ⊥ AC and BM ⊥ AC.

Proof:

In ΔOND and ΔOMB,

⇒ OD = OB {Given}

⇒ ∠DON = ∠BON {vertically opposite angles}

⇒ ∠OND = ∠OMB {Each = 90°}

⇒ ΔOND ≅ ΔOMB

DN = BM {c.p.c.t}

ON = OM

In ΔDCN and ΔBAM,

DN = BM

DC = AB {given}

⇒ ∠CND = ∠AMB {Each = 90°}

⇒ ΔDCN ≅ ΔBAM {RHS congruence}

⇒ CN = AM {c.p.c.t}

ON = OM

⇒ CN + ON = AM + OM

⇒ OC = OA

In ΔDOC and ΔBOA,

⇒ OD = OB {given}

⇒ DC = AB {given}

⇒ OC = OA {proved above}

⇒ ΔDOC ≅ ΔBOA

⇒ ar(ΔDOC) ≅ ar(ΔBOA)

⇒ ar(ΔDOC) = ar(ΔAOB)

(ii) Adding ar(ΔOCB) on both sides, we get

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

⇒ ar(ΔDCB) = ar(ΔACB)

(iii) In ABCD,

OB = OD {given}

OC = OA {proved}

The diagonals AC and BD of Quad.ABCD bisect each other.

⇒ ABCD is a parallelogram.

Hope it helps!