Advertisements
Advertisements
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.]
Advertisements
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
D and E are points on sides AB and AC respectively of ΔABC such that
ar (DBC) = ar (EBC). Prove that DE || BC.
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure). Show that
ar (ABCD) = ar (PBQR).
[Hint: Join AC and PQ. Now compare area (ACQ) and area (APQ)]
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
(i) ar (BDE) = 1/4 ar (ABC)
(ii) ar (BDE) = 1/2 ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = 1/8 ar (AFC)
[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint : From A and C, draw perpendiculars to BD.]
In a ΔABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point
of AP. Prove that :
(1) ar (Δ PBQ) = ar (Δ ARC)
(2) ar (Δ PRQ) =`1/2`ar (Δ ARC)
(3) ar (Δ RQC) =`3/8` ar (Δ ABC) .
X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See figure). Prove that ar (LZY) = ar (MZYX)
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔABD)
A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)
