In the given figure, ABCD is a parallelogram, O is any point on AC, PQ||AB and LM||AD. Prove that: ar(||gm DLOP)=ar(||gm BMOQ).

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In the figure, ABCD is a ||gm . O is any point on AC. PQ || AB and LM || AD. Prove that ar (||gm DLOP) = ar(||gm BMOQ).

ar (∆ O A B) + ar (∆ O C D) = \frac{1}{2} ar (| | gm A B C D),

ar (∆ O A D) + ar (∆ O B C) = \frac{1}{2} ar (| | gm A B C D) .

(△ E B D)

=

(△ E D C) .

(△ A B E)

=

(△ A C E) .

In the given figure, ABCD is a parallelogram. P and Q are any two points in AB and BC respectively. Area of $Δ$ ADQ= ___

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