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Condition for Two Lines to Be Parallel

In Fig. diago...

Question

In Fig. diagonals $A C$ and $B D$ of quadrilateral $A B C D$ intersect at $O$ such that $O B = O D$ . If $A B = C D$ , then show that :
i) ar $(△ D O C) =$ ar( $△ A O B)$
ii) ar $(△ D C B) =$ ar $(△ A C B)$
iii) $D A ∥ C B$ or $A B C D$ is parallelogram.

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Solution

(i) $a r (D O C) = a r (A O B)$

Given: A quadrilateral ABCD where $O B = O D$ & $A B = C D .$

To prove: $a r (D O C) = a r (A O B)$

Proof: Let us draw $O P ⊥ A C$ and $B Q ⊥ A C$ :

In $Δ$ DOP and $Δ$ BOQ,

$∠ D P O = ∠ B Q O$ (Both $90^{\circ}$ )

$∠ D O P = ∠ B O Q$ (vertically opposite angles)

$O D = O B$ (Given)

$Δ D O P ≅ Δ B O Q$ (ASS congruence rule)

$∴$ $D P = B Q$ (CPT) ……… $(1)$

and ar( $D O P$ )=ar( $B O Q$ ) $\dots \dots \dots \dots \dots . .$ (2)$
(Area of congruent triangles are equal)

In $Δ C O P$ and $Δ A B Q,$

$∠ C P D = ∠ A Q B$ (Both $90^{o}$ )
$C P = A B$ (Given)
$D P = B Q$ (from $(1)$ )
$∴ Δ C D P ≅ Δ A B Q$ (RHS congruence rule)
$\Rightarrow a r (C D P) = a r (A B Q)$ ………… $(3)$
(Area of congruent triangles are equal)

Adding $(2)$ & $(3)$ :

$a r (D O P) + a r (C P D) = a r (B O Q) + a r (A B Q)$

$a r (D O C) = a r (A O B)$

Hence proved.

(ii) $a r (D C B) = a r (A C B)$

In part (i) we proved that
$a r (D O C) = A r (A O B)$

Adding $a r (O C B)$ both sides
$a r (D O C) + a r (O C B) = a r (A O B) + a r (O C B)$
$\Rightarrow a r (D C B) = a r (A C B)$ .

(iii) In part (ii) we proved

$a r (Δ D B C) = a r (Δ A B C)$

We know that two trinagles having same base & equal areas, lie between same parallels.

Here $Δ D B C$ & $Δ A B C$ are on the same base BC & are equal in area.

So, these triangles lie between the same parallels $D A$ and $C B$ .
$∴ D A | | C B$ .

Now,

In $Δ D O A$ & $Δ B O C$

$∠ D O A = ∠ B O C$ (vertically opposite angles)
$∠ D A O = ∠ B C O$
( $D A | | B C, A C$ traversal, alternate angles equal)
$O B = O B$ (Given)

$∴ Δ D O A ≅ Δ B O C$ (ASA congruency)
$D A = B C$ (CPCT)

So, in $A B C D$ , since $D A | | C B$ and $D A = C B$ .

One pair of opposite sides of quadrilateral $A B C D$ is equal and parallel.
$∴$ $A B C D$ is a parallelogram.
Hence proved.

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A C

B D

A B C D

O

O B = O D

A B = C D

a r (D O C) = a r (A O B)

a r (D C B) = a r (A C B)

D A ∥ C B

A B C D

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.]

(Δ D O C)

(Δ A O B)

(Δ D C B)

(Δ A C B)

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