$A B C D$ is a parallelogram, a line through $A$ cuts $D C$ at point $P$ and $B C$ produced at $Q$ .
Prove that triangle $B C P$ is equal in area to triangle $D P Q$ .

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Solution

Given: In parallelogram $A B C D,$ $A B ∥ C D$ and $A D ∥ B C$

Now, join a line from $A$ to $C$

We get $△ A P C$
Here, $△ A C P$ and $△ B C P$ lie on the same base $C P$ and between the same parallels $A B$ and $C P$

Hence $A r e a (△ A C P) = A r e a (△ B C P)$ $. . . (1)$

Also, $△ A D Q$ and $△ A D C$ lie on the same base $A D$ and lie between the same parallels $A D$ and $B C$

$∴ A r e a (△ A D Q) = A r e a (△ A D C)$ $. . . (2)$
$\Rightarrow A r e a (△ A D Q) - A r e a (△ A D P) = A r e a (△ A D C) - A r e a (△ A D P)$
$\Rightarrow A r e a (△ A P C) = A r e a (△ D P Q)$ $. . . (3)$
From $(1)$ and $(3),$ we get
$\Rightarrow A r e a (△ B P C) = A r e a (△ D P Q)$

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ABCD is a parallelogram, a line through A cuts DC at point P and BC produced at Q .Prove that triangle BCP is equal in area to triangle DPQ.

$A B C D$ is a parallelogram, a line through $A$ cuts $D C$ at point $P$ and $B C$ produced at $Q$ .
Prove that triangle $B C P$ is equal in area to triangle $D P Q$ .