In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
Solution:

Since $\triangle A B Q$ and $\triangle P B Q$ lie on the same base $B Q$ and are between the same parallels $A P$ and $B Q$,

$\therefore$ Area $(\Delta \mathrm{ABQ})=$ Area $(\triangle \mathrm{PBQ}) \ldots(1)$

Again, $\triangle B C Q$ and $\triangle B R Q$ lie on the same base $B Q$ and are between the same parallels $B Q$ and $C R$.

$\therefore$ Area $(\Delta B C Q)=$ Area $(\Delta B R Q) \ldots(2)$

On adding equations (1) and (2), we obtain

Area $(\Delta \mathrm{ABQ})+$ Area $(\Delta \mathrm{BCQ})=$ Area $(\Delta \mathrm{PBQ})+$ Area $(\Delta \mathrm{BRQ})$

$\Rightarrow$ Area $(\triangle A Q C)=$ Area $(\triangle P B R)$
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