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Question Number 177046 by Ar Brandon last updated on 30/Sep/22

Answered by som(math1967) last updated on 30/Sep/22

$$Area(△ADE+△BEC)=\frac{1}{2}AreaABCD$$$$=\frac{x}{2}\left(let\right)$$$$Area△AEB=\frac{1}{2}ABCD=\frac{x}{2}$$$$Area(ADEP+△ADE)$$$$=△BEC+△ADE=\frac{x}{2}$$$$area△APE=Area(ADEP+△ADE)$$$$=\frac{x}{2}$$$$\therefore ar△APE=ar△AEB$$$$\Rightarrow \frac{ar△APE}{ar△AEB}=\frac{1}{1}$$$$\Rightarrow \frac{PE}{EB}=\frac{1}{1}$$$$$$

Commented by som(math1967) last updated on 30/Sep/22

Commented by Ar Brandon last updated on 30/Sep/22

Thank you Sir

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