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Distances between Special Points in a Triangle

In the figure...

Question

In the figure, diagonals $A C$ and $B D$ of a quadrilateral $A B C D$ intersect at $O$ such that $O B = O D$ , show that $a r (Δ D O C) = a r (Δ A O B)$

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Solution

$B O = O D$
Since $B O = O D$ is the mid point
we know thta $a$ mid point of a $△$ divides it into two $△$ of equal areas
$C O$ is medpoint of $△ B C D$
i.e., $a r (△ C O D) = a r (△ C O B) - - - (1)$
$A O$ is a medpoint of $△ A B D$
i.e. $a r (△ A O D) = a r (△ A O B) - - - (2)$
from $(I)$ & $(I I)$ we have
$a r (△ C O D) + a r (△ A O D) = a r (△ C O B) + a r (△ A O B)$
$a r (△ A D C) = a r (△ A B C)$

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