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Question
In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC.]
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Solution
Let us draw DN ⊥ AC and BM ⊥ AC.
(i) In ΔDON and ΔBOM,
∠DNO = ∠BMO (By construction)
∠DON = ∠BOM (Vertically opposite angles)
OD = OB (Given)
By AAS congruence rule,
ΔDON ≅ ΔBOM
∴ DN = BM ... (1)
We know that congruent triangles have equal areas.
∴ Area (ΔDON) = Area (ΔBOM) ... (2)
In ΔDNC and ΔBMA,
∠DNC = ∠BMA (By construction)
CD = AB (Given)
DN = BM [Using equation (1)]
∴ ΔDNC ≅ ΔBMA (RHS congruence rule)
⇒ Area (ΔDNC) = Area (ΔBMA) ... (3)
On adding equations (2) and (3), we obtain
Area (ΔDON) + Area (ΔDNC) = Area (ΔBOM) + Area (ΔBMA)
Therefore, Area (ΔDOC) = Area (ΔAOB)
(ii) We obtained,
Area (ΔDOC) = Area (ΔAOB)
⇒ Area (ΔDOC) + Area (ΔOCB) = Area (ΔAOB) + Area (ΔOCB)
(Adding Area (ΔOCB) to both sides)
⇒ Area (ΔDCB) = Area (ΔACB)
(iii) We obtained,
Area (ΔDCB) = Area (ΔACB)
If two triangles have the same base and equal areas, then these will lie between the same parallels.
∴ DA || CB ... (4)
In ΔDOA and ΔBOC,
∠DOA = ∠BOC (Vertically opposite angles)
OD = OB (Given)
∠ODA = ∠OBC (Alternate opposite angles)
By ASA congruence rule,
ΔDOA ≅ ΔBOC
∴ DA = BC ... (5)
In quadrilateral ABCD, one pair of opposite sides is equal and parallel (AD = BC)
Therefore, ABCD is a parallelogram.
RELATED QUESTIONS
In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE)
XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that
ar (ABE) = ar (ACF)
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) ar(PRQ) = 1/2 ar(ARC)
(ii) ar(RQC) = 3/8 ar(ABC)
(iii) ar(PBQ) = ar(ARC)
In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:-
(i) ΔMBC ≅ ΔABD
(ii) ar (BYXD) = 2 ar(MBC)
(iii) ar (BYXD) = ar(ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar (CYXE) = ar(ACFG)
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)
Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.
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(3) ar (ΔABM) ar (ΔACL)
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