Question

Solve this........... ​

Answer 1

Answer:

hloo mate..

here's ur answer...

Given: OB=OD and AB=CD

Construction: DE perpendicular on AC and BF perpendicular on AC

Proof:

(i) In ΔDOE and ΔBOF,

∠DEO = ∠BFO (Perpendiculars)

∠DOE = ∠BOF (Vertically opposite angles)

OD = OB (Given)

∴ ΔDOE ≅ ΔBOF by AAS congruence condition.

∴ DE = BF (By CPCT) — (i)

also, ar(ΔDOE) = ar(ΔBOF) (Congruent triangles) — (ii)

Now,

In ΔDEC and ΔBFA,

∠DEC = ∠BFA (Perpendiculars)

CD = AB (Given)

DE = BF (From i)

∴ ΔDEC ≅ ΔBFA by RHS congruence condition.

∴ ar(ΔDEC) = ar(ΔBFA) (Congruent triangles) —(iii)

Adding (ii) and (iii),

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

⇒ ar (DOC) = ar (AOB)

(ii) ar(ΔDOC) = ar(ΔAOB)

Adding ar(ΔOCB) in LHS and RHS, we get,

⇒ar(ΔDOC)+ar(ΔOCB)=ar(ΔAOB)+ar(ΔOCB)

⇒ ar(ΔDCB) = ar(ΔACB)

(iii) When two triangles have same base and equal areas, the triangles will be in between the same parallel lines

ar(ΔDCB) = ar(ΔACB)

DA || BC — (iv)

For quadrilateral ABCD, one pair of opposite sides are equal (AB = CD) and other pair of opposite sides are parallel.

∴ ABCD is parallelogram.

hope it helps..

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Answer 2

Step-by-step explanation:

( hint: from D and B, draw perpendicular to AC)

OB =OD

AB =CD

let DM and BN are altitude on AC

angle DOM=angleBON ( vertically opposite angle)

OB=OD ( given)

angleDMO=angleBNO ( 90 degree)

angle DOM=angle BON ( by ASA)

DM=BN (by cpct)

area (DOM) =area (BON) ( by cpct) ( equation 1 )

In triangle DMC and triangle BNA

CD = AB ( given)

angle DMC=angleBNA ( 90 degree)

DM=BN (proved)

triangle DMC congurent triangle BNA ( by SAS)

area (DMC) =area)(BNA) ( by cpct) (equation 2)

adding eqn 1. and 2.

area (DOM) + area (DCM) = area (BON) + area (BNA)

area (DOC) =area (AOB) ( equation 3 )

Adding area (BOC) on both side in equation 3.

area (DOC) + area (BOC)= area (AOB) + area (BOC)

area (DCB) = area (ACB) (equation 4)

triangle DOC congurent triangle AOB (proved)

angle CDO congurent angle OBA ( by cpct)

But angle CDO and angle OBA are the pairs of alternate angle form by parallel lines AB and CD which are AB = CD

quadrilateral ABCD is a parallelogram

therefore, AD||BC

I hope it helps u