Question.
In the given figure, $A P\|B Q\| C R .$ Prove that $\operatorname{ar}(A Q C)=\operatorname{ar}(P B R) .$
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 $(\triangle A B Q)=$ Area $(\triangle P B Q) \ldots(1)$
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 $(\triangle P B R)$
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 $(\triangle A B Q)=$ Area $(\triangle P B Q) \ldots(1)$
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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