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Given

: Two triangles ABC and DBC are on the same base BC and between same parallels EF and BC.

To Prove : ar(ΔABC) = ar(ΔDBC)

Construction : Through B, draw BE || AC intersecting line AD in E and through C draw CF || BD intersecting the line DA in F.

Proof : EACB and DFCB are parallelograms (Since two pairs of opposite sides are parallel)

Also ||gm EACB and ||gm DFCB are on the same base BC and between same parallels EF and BC.

∴ ar(||gm EACB)= ar(||gm DFCB)...(i)

Now AB is the diagonal of ||gm EACB

∴ ar(ΔEAB) = ar(ΔABC)

∴ ar(ΔABC) = (1/2)ar(||gm EACB) .....(ii)

Similarly,

ar(ΔDBC) = (1/2) ar(||gm DFCB)...(iii)

From equations (i), (ii) and (iii), we get

ar(Δ ABC) = ar(Δ DBC)

See the figure

Given

: Two triangles ABC and DBC are on the same base BC and between same parallels EF and BC.

To Prove : ar(ΔABC) = ar(ΔDBC)

Construction : Through B, draw BE || AC intersecting line AD in E and through C draw CF || BD intersecting the line DA in F.

Proof : EACB and DFCB are parallelograms (Since two pairs of opposite sides are parallel)

Also ||gm EACB and ||gm DFCB are on the same base BC and between same parallels EF and BC.

∴ ar(||gm EACB)= ar(||gm DFCB)...(i)

Now AB is the diagonal of ||gm EACB

∴ ar(ΔEAB) = ar(ΔABC)

∴ ar(ΔABC) = (1/2)ar(||gm EACB) .....(ii)

Similarly,

ar(ΔDBC) = (1/2) ar(||gm DFCB)...(iii)

From equations (i), (ii) and (iii), we get

ar(Δ ABC) = ar(Δ DBC)