1
Question
In the given figure, the diagonals AC and BD intersect at point O. If OB = OD and AB//DC, show that :Area(ΔDCB)=Area(ΔACB).
Area(ΔDCB)=Area(ΔACB).
Open in App
Solution
Let us draw DN⊥AC and BM⊥AC as shown in above figure.
In triangles DON and BMO,
⊥DNO=⊥BMO ( by construction)
⊥DON=⊥BOM ( vertically opposite angles)
OD=OB (given)
By AAS congruence rule,
ΔDON⊥ΔBOM
DN=BM.............( eqn 1)
We know that congruent triangles have equal areas, thus, Ar(ΔDON)=Ar(ΔBOM).
In triangles DNC and BMA,
⊥DNC=⊥BMA ( by construction)
CD=AB (given)
DN=BM ( using eqn 1)
ΔDNC⊥ΔBMA ( RHS congruence rule)
Therefore, Ar(ΔDOC)=Ar(ΔAOB).
Adding areas on both sides, we get
Ar(ΔDNC)+Ar(ΔOCB)=Ar(ΔAOB)+Ar(ΔOCB)
⇒Ar(ΔDCB)=Ar(ΔACB)
Hence proved.
In triangles DON and BMO,
⊥DNO=⊥BMO ( by construction)
⊥DON=⊥BOM ( vertically opposite angles)
OD=OB (given)
By AAS congruence rule,
ΔDON⊥ΔBOM
DN=BM.............( eqn 1)
We know that congruent triangles have equal areas, thus, Ar(ΔDON)=Ar(ΔBOM).
In triangles DNC and BMA,
⊥DNC=⊥BMA ( by construction)
CD=AB (given)
DN=BM ( using eqn 1)
ΔDNC⊥ΔBMA ( RHS congruence rule)
Therefore, Ar(ΔDOC)=Ar(ΔAOB).
Adding areas on both sides, we get
Ar(ΔDNC)+Ar(ΔOCB)=Ar(ΔAOB)+Ar(ΔOCB)
⇒Ar(ΔDCB)=Ar(ΔACB)
Hence proved.
Suggest Corrections
0
Similar questions
