The given figure shows a parallelogram $A B C D . B C$ is extended to the point $P . P C$ is joined intersecting $A C$ to $O$ .
Show that: $D O \times C P = O P \times B C$ .

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Solution

$A B C D$ is a parallelogram.

$A D ∥ B C$ [ Opposite sides of parallelogram are parallel ]

$∴$ $A D ∥ C P$ [ Since, $C P$ is extended part of $B C$ ]

In $△ A D O$ and $△ C P O$

$A D ∥ C P$ and $A C$ is transversal.

$\Rightarrow$ $∠ D A O = ∠ O C P$ [ Alternate angles ]

$A D ∥ C P$ and $D P$ is transversal.

$\Rightarrow$ $∠ A D O = ∠ O P C$ [ Alternate angles ]

$∴$ $△ A D O \sim △ C P O$

$\Rightarrow$ $\frac{A D}{C P} = \frac{D O}{O P}$

$\Rightarrow$ $A D \times O P = D O \times C P$

$A D = B C$ [ Opposite sides of a parallelogram are equal ]

$\Rightarrow$ $B C \times O P = D O \times C P$

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The given figure shows a parallelogram ABCD.BC is extended to the point P.PC is joined intersecting AC to O.Show that: DO×CP=OP×BC.

The given figure shows a parallelogram $A B C D . B C$ is extended to the point $P . P C$ is joined intersecting $A C$ to $O$ .
Show that: $D O \times C P = O P \times B C$ .