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Question Number 189070 by Rupesh123 last updated on 11/Mar/23

Answered by a.lgnaoui last updated on 12/Mar/23

$$consideronsPointintersectionde$$$$AE,BFetAD$$$$\Rightarrow \{{}_{\measuredangle MPF=\measuredangle BPN}^{\measuredangle BPM=\measuredangle NPE}$$$$suposons\frac{BM}{MF}=\frac{BN}{NE}=1$$$$akorsBM=MFBN=NE$$$$MN\mid \mid BCMN=x+y$$$$△BFC△BECsemblables$$$$(memecoteBC$$$$△BFC\frac{BM}{MF}=\frac{BC}{x}\left(1\right)$$$$△BEC\frac{CN}{NE}=\frac{BC}{y}\left(2\right)$$$$d^{\prime }autrepart:△ABDet△ADC$$$$\frac{AP}{PD}=\frac{AM}{BM}=\frac{AN}{NB}\Rightarrow \frac{x}{BD}=\frac{y}{CD}\left(3\right)$$$$$$$$\frac{\left(1\right)}{\left(2\right)}=\frac{y}{x}=\frac{BM}{MF}\times \frac{NE}{NC}$$$$\frac{BM}{MF}\times \frac{x}{y}=\frac{NC}{NE}\left(4\right)$$$$\left(3\right)\frac{x}{y}=\frac{BD}{CD}\left(5\right)$$$$\frac{\left(4\right)}{\left(5\right)}\Rightarrow \frac{BM}{MF}=\frac{NC}{NE}\times \frac{CD}{BD}$$$$commeBM=MF\Rightarrow$$$$BD=CD$$$$donc\frac{BM}{MF}=\frac{x}{y}=\frac{NC}{NE}=\frac{NE}{NC}=\frac{BD}{CD}$$$$\Rightarrow \bullet BM=MF\Rightarrow \{\begin{array}{l}NE=NC\\ x=y\end{array}$$$$et(EF\mid \mid MN\mid \mid BC)$$

Commented by a.lgnaoui last updated on 12/Mar/23

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