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State true or false.

$A B C D$ is a parallelogram whose diagonals intersect at $O$ is a parallelogram whose diagonals intersect at . If $P$ . If is any point on $B O$ is any point on , then $a r e a Δ A B P = a r e a Δ C B P$ , then .

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True

False

answer is A.

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Detailed Solution

Given that,

A B C D

is a parallelogram whose diagonals intersect at

O

and

P

is any point on

B O

.
We have to prove that,

a r e a Δ A B P = a r e a Δ C B P

The figure is given below,
Question Image

Diagonals of the parallelogram bisect each other.
AO = OC and BO = OD.
So,

O

is the midpoint of

B D

.
We know that, the median divides the triangle into two equal areas.
As

B O

is the median of

Δ B A C

a r e a Δ B O A = a r e a Δ B O C

. . . . . . (1)

P O

is the median of

Δ P A C

a r e a Δ P O A = a r e a Δ P O C

. . . . . . (2)
Subtracting (2) from (1),

a r e a (Δ B O A) − a r e a (Δ P O A) = a r e a (Δ B O C) − a r e a (Δ P O C)

a r e a (Δ A B P) = a r e a (Δ C B P)

Hence, the statement is true.
Therefore, option 1 is correct.

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