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Question-177046




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