wangsacl wrote: For every triangle $ ABC$, let $ D,E,F$ be a point located on segment $ BC,CA,AB$, respectively. Let $ P$ be the intersection of $ AD$ and $ EF$. Prove that: \[ \frac {AB}{AF}\times DC + \frac {AC}{AE}\times DB = \frac {AD}{AP}\times BC\] It is nice but easy!. In the case $ AB<AC$ Let $ BI$;$ CJ$ be parallel to $ EF$ where $ I;J$ belong to $ AD$. Thales Theorem shows us $ \frac{AB}{AF}=\frac{AI}{AP}$;$ \frac{AC}{AE}=\frac{AJ}{AP}$;$ \frac{ID}{DB}=\frac{JD}{DC}=\frac{IJ}{BC}$. (1) Notice that $ AI=AD-ID=AD-\frac{DB.IJ}{BC}$ thus $ \frac{AI.DC}{AP}=\frac{AD.DC}{AP}-\frac{IJ.DB.DC}{BC.AP}$ A same thing is $ \frac{AJ.DB}{AP}=\frac{AD.DB}{AP}+\frac{IJ.DB.DC}{BC.AP}$. Now , it is immediate that $ \frac{AI.DC}{AP}+\frac{AJ.DB}{AP}=\frac {AD}{AP}\times BC$ since the fact $ DC+DB=BC$. Now, the proof can be completed after we look at again (1)

Yup. The official solution only uses ratios in length and area, which is very-very beautiful (and algebraic )

Here is my solution $ \frac{AD}{AP}.\frac{AB}{AF}.\frac{DC}{BC}+\frac{AD}{AP}.\frac{AC}{AE}.\frac{DB}{BC}=\frac{[ABD]}{[AFP]}.\frac{[ADC]}{[ABC]}+\frac{[ADC]}{[APE]}.\frac{[ABD]}{[ABC]}=\frac{[ABD].[ADC]}{[ABC]}(\frac{1}{[AFP]}+\frac{1}{[APE]})=\frac{[ABD].[ADC]}{[AFP].[APE]}.\frac{[AFE]}{[ABC]}=\frac{AB.AD}{AF.AP}.\frac{AD.AC}{AP.AE}.\frac{AF.AE}{AB.AC}=\frac{AD^2}{AP^2}$ So, $ \frac{AD}{AP}.\frac{AB}{AF}.\frac{DC}{BC}+\frac{AD}{AP}.\frac{AC}{AE}.\frac{DB}{BC}=\frac{AD^2}{AP^2}$, which is equivalent with the problem $ QED$

it's (almost) Cristea's Theorem: let ABC be a scalene triangle and D a point on the segment BC. M belongs to the segment AD. E and F belong to the segment AB and AC ,respectively. Then EF passes through M if and only if: DC(EB/EA) + BD(FC/FA) = BC(MD/MA)

BTW, that Cristea's Theorem is not well-known to me. But this problem is just an old result from http://www.mathlinks.ro/viewtopic.php?p=1143691#1143691

A simple proof is using vector. We have $ BD. \vec{AC}+DC.\vec{AB}=BC.\vec{AD}$ $ \Rightarrow \frac{AC}{AE}.BD.\vec{AE}+\frac{AB}{AF}.DC.\vec{AF}=\frac{AD}{AP}.BC.\vec{AP}$ or $ x.\vec{AE}+y.\vec{AF}=z.\vec{AP}$ On the other side, $ FP.\vec{AE}+EP.\vec{AF}=EF.\vec{AP}$ So $ \frac{x}{FP}=\frac{y}{EP}=\frac{z}{EF}=\frac{x+y}{FP+EP}=\frac{x+y}{EF}$ $ \Rightarrow x+y=z$ (QED)

wangsacl wrote: For every triangle $ ABC$, let $ D,E,F$ be a point located on segment $ BC,CA,AB$, respectively. Let $ P$ be the intersection of $ AD$ and $ EF$. Prove that: \[ \frac {AB}{AF}\times DC + \frac {AC}{AE}\times DB = \frac {AD}{AP}\times BC\] I am giving a solution using Menelaus Theorem. Construct $ F',E'$ on $ AB, AC$ resp. such that $ F'E' \parallel BC$ Now Menelaus on $ \triangle AF'E'$ and transversal $ FPE$ gives, \begin{align*} & \frac{AF}{FF'}\cdot \frac{F'P}{PE'} \cdot \frac{EE'}{EA}=1 \\ \iff & \frac{PE'}{PF'}=\frac{AF}{FF'} \cdot \frac{EE'}{EA} \\ \iff & \frac{DC}{DB}\cdot \frac{AB}{AC}=\frac{EE'}{FF'}\cdot \frac{AF}{AE} \cdot \frac{AF'}{AE'}=\frac{\frac{AE'-AE}{AE\times AE'}}{\frac{AF-AF'}{AF\times AF'}}\\ \iff & AB\times DC \left[\frac{1}{AF'}-\frac{1}{AF}\right]=AC\times DB \left[\frac{1}{AE}-\frac{1}{AE'}\right] \\ \iff & AB\times DC \times \frac{1}{AF'}+AC\times DB \times \frac{1}{AE'} = AB\times DC \times \frac{1}{AF}+AC\times DB \frac{1}{AE} \end{align*} But \begin{align*} \frac{AD}{AP}\times BC=& \frac{AD}{AP}\times (BD+DC)=\frac{AB}{AF'}\times DC+\frac{AC}{AE'}\times BD\\ =&\frac {AB}{AF}\times DC + \frac {AC}{AE}\times DB \end{align*}

Attachments: