Question

please solve it fast​

Answer 1

Answer:

This can be proved by theorem that triangle on same base and between same parallel are equal in areas .

Step-by-step explanation:

To prove - DG║FC

Construction - Join GC and DF

Proof - Here, area of parallelogram ABCD = area of parallelogram AGFE

=> area of parallelogram AGOD + area of parallelogram GBCO = area of parallelogram AGOD + area of parallelogram DOFE

=>  area of parallelogram GBCO =  area of parallelogram DOFE

=>  1/2 * area of parallelogram GBCO =  1/2 * area of parallelogram DOFE

=> area of ΔGOC = area of ΔDOF

Adding area ΔDOG on both sides

=>  area of ΔGOC + area ΔDOG = area of ΔDOF + area ΔDOG

=> area ΔDGC = area ΔDGF

As, ΔDGC and ΔDGF are on same base DG and have equal areas .

So, ΔDGC and ΔDGF lie between the same parallels DG and FC

Hence , DG║FC

Proved .

Answer 2

Answer:

Proved

Step-by-step explanation:

to Prove DG ║ FC

we need to show that

OG/OD  = OF/OC

Area of ABCD = OG * AB

Area of AGFE = OD * AG

both areas are equal

OG * AB = OD * AG

=> AG = (OG * AB)/OD  - Eq 1

OG/OD  = OF/OC

=> OG/OD  = (AG - OG)/(AB - OD)

Putting value of AG from Eq 1

=> OG/OD = ((OG * AB)/OD - OG ) / (AB - OD)

=> OG/OD = ((OG * AB) - (OD * OG) ) / (OD *(AB - OD))

=> OG/OD = (OG * (AB - OD) ) / (OD *(AB - OD))

Cancelling AB - OD

=> OG/OD = OG/OD

=> LHS  = RHS

QED