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Question-189070




Question Number 189070 by Rupesh123 last updated on 11/Mar/23
Answered by a.lgnaoui last updated on 12/Mar/23
considerons Point intersection de   AE ,BF et AD   ⇒{_(∡MPF=∡BPN) ^(∡BPM=∡NPE)   suposons  ((BM)/(MF))=((BN)/(NE))=1  akors    BM=MF   BN=NE  MN∣∣ BC   MN=x+y  △BFC   △BEC semblables  (meme cote BC  △BFC  ((BM)/(MF))=((BC)/x)    (1)  △BEC  ((CN)/(NE))=((BC)/y)      (2)  d′autre part :△ABDet  △ADC   ((AP)/(PD))=((AM)/(BM))=((AN)/(NB))⇒(x/(BD))=(y/(CD))  (3)    (((1))/((2)))=(y/x)=((BM)/(MF))×((NE)/(NC))  ((BM)/(MF))×(x/y)=((NC)/(NE))     (4)  (3)    (x/y)=((BD)/(CD))       (5)  (((4))/((5)))⇒((BM)/(MF))=((NC)/(NE))×((CD)/(BD))  comme BM=MF⇒                    BD=CD  donc     ((BM)/(MF))=(x/y)=((NC)/(NE))=((NE)/(NC))=((BD)/(CD))  ⇒ •BM= MF⇒ { ((NE=NC)),((     x=y)) :}               et (EF∣∣MN∣∣BC)
$${considerons}\:{Point}\:{intersection}\:{de}\: \\ $$$${AE}\:,{BF}\:{et}\:{AD}\: \\ $$$$\Rightarrow\left\{_{\measuredangle{MPF}=\measuredangle{BPN}} ^{\measuredangle{BPM}=\measuredangle{NPE}} \right. \\ $$$${suposons}\:\:\frac{{BM}}{{MF}}=\frac{{BN}}{{NE}}=\mathrm{1} \\ $$$${akors}\:\:\:\:{BM}={MF}\:\:\:{BN}={NE} \\ $$$${MN}\mid\mid\:{BC}\:\:\:{MN}={x}+{y} \\ $$$$\bigtriangleup{BFC}\:\:\:\bigtriangleup{BEC}\:{semblables} \\ $$$$\left({meme}\:{cote}\:{BC}\right. \\ $$$$\bigtriangleup{BFC}\:\:\frac{{BM}}{{MF}}=\frac{{BC}}{{x}}\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\bigtriangleup{BEC}\:\:\frac{{CN}}{{NE}}=\frac{{BC}}{{y}}\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$${d}'{autre}\:{part}\::\bigtriangleup{ABDet}\:\:\bigtriangleup{ADC}\: \\ $$$$\frac{{AP}}{{PD}}=\frac{{AM}}{{BM}}=\frac{{AN}}{{NB}}\Rightarrow\frac{{x}}{{BD}}=\frac{{y}}{{CD}}\:\:\left(\mathrm{3}\right) \\ $$$$ \\ $$$$\frac{\left(\mathrm{1}\right)}{\left(\mathrm{2}\right)}=\frac{{y}}{{x}}=\frac{{BM}}{{MF}}×\frac{{NE}}{{NC}} \\ $$$$\frac{{BM}}{{MF}}×\frac{{x}}{{y}}=\frac{{NC}}{{NE}}\:\:\:\:\:\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{3}\right)\:\:\:\:\frac{{x}}{{y}}=\frac{{BD}}{{CD}}\:\:\:\:\:\:\:\left(\mathrm{5}\right) \\ $$$$\frac{\left(\mathrm{4}\right)}{\left(\mathrm{5}\right)}\Rightarrow\frac{{BM}}{{MF}}=\frac{{NC}}{{NE}}×\frac{{CD}}{{BD}} \\ $$$${comme}\:{BM}={MF}\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{BD}={CD} \\ $$$${donc}\:\:\:\:\:\frac{{BM}}{{MF}}=\frac{{x}}{{y}}=\frac{{NC}}{{NE}}=\frac{{NE}}{{NC}}=\frac{{BD}}{{CD}} \\ $$$$\Rightarrow\:\bullet{BM}=\:{MF}\Rightarrow\begin{cases}{{NE}={NC}}\\{\:\:\:\:\:{x}={y}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{et}\:\left({EF}\mid\mid{MN}\mid\mid{BC}\right)\:\: \\ $$
Commented by a.lgnaoui last updated on 12/Mar/23

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