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orl wrote: Prove that $E, E_1F_1, PD_1$ are concurrent. I don't understand.

A hardy attempt to reconstruct the correct problem statement: Problem. Let D, E, F be points on the sides BC, CA, AB of a triangle ABC such that EF || BC. Let $D_1$ be a point on the side BC, and let the parallels to the lines DE and DF through the point $D_1$ meet the lines AC and AB at the points $E_1$ and $F_1$, respectively. Let the parallels to the lines DF and DE through the points B and C intersect at a point P. Prove that the lines EF, $E_1F_1$ and $PD_1$ are concurrent if and only if the point $D_1$ is the midpoint of the segment BC.

Solution. Since both of the lines BP and $D_1F_1$ were constructed as parallels to the line DF, we get a triple of parallel lines: $BP\parallel D_1F_1\parallel DF$. Similarly, $CP\parallel D_1E_1\parallel DE$. Let the lines BP and CP meet the line $E_1F_1$ at the points U and V, respectively. Let the lines $PD_1$ and $E_1F_1$ intersect at R. Then, since $BP\parallel D_1F_1$, by Thales we have $\frac{UR}{F_1R}=\frac{PU}{D_1F_1}=\frac{PR}{D_1R}$, and since $CP\parallel D_1E_1$, Thales yields $\frac{VR}{E_1R}=\frac{VP}{E_1D_1}=\frac{PR}{D_1R}$. Thus, $\frac{UR}{F_1R}=\frac{PU}{D_1F_1}=\frac{VR}{E_1R}=\frac{VP}{E_1D_1}$. This equation splits into $\frac{UR}{F_1R}=\frac{VR}{E_1R}$, what is equivalent to $\frac{UR}{F_1R}=\frac{RV}{RE_1}$ or to $\frac{UR}{RV}=\frac{F_1R}{RE_1}$, and into $\frac{PU}{D_1F_1}=\frac{VP}{E_1D_1}$, what becomes $\frac{VP}{PU}=\frac{E_1D_1}{D_1F_1}$. Since BC || FE, CP || ED and PB || DF (these are just the relations EF || BC, CP || DE and BP || DF, slightly rewritten), the triangles BCP and FED are homothetic. Thus, $\frac{BP}{PC}=\frac{FD}{DE}$. In other words, $\frac{BP}{PC}=\frac{DF}{ED}$. Now we will use a fact which, if not known, is a nice little exercise for the reader (it should appear in good books, alas it doesn't...): Menelaos theorem for quadrilaterals. Let ABCD be a quadrilateral, and let X, Y, Z, W be four collinear points on its sidelines AB, BC, CD, DA, respectively. Then, $\frac{AX}{XB}\cdot\frac{BY}{YC}\cdot\frac{CZ}{ZD}\cdot\frac{DW}{WA}=1$. Applying the Menelaos theorem for quadrilaterals to the quadrilateral UVCB with the collinear points R, P, $D_1$, P on its sidelines UV, VC, CB, BU, respectively, we get $\frac{UR}{RV}\cdot\frac{VP}{PC}\cdot\frac{CD_1}{D_1B}\cdot\frac{BP}{PU}=1$. In other words: $\frac{UR}{RV}\cdot\frac{VP}{PU}\cdot\frac{BP}{PC}\cdot\frac{CD_1}{D_1B}=1$. Using the relations $\frac{UR}{RV}=\frac{F_1R}{RE_1}$, $\frac{VP}{PU}=\frac{E_1D_1}{D_1F_1}$ and $\frac{BP}{PC}=\frac{DF}{ED}$, this becomes $\frac{F_1R}{RE_1}\cdot\frac{E_1D_1}{D_1F_1}\cdot\frac{DF}{ED}\cdot\frac{CD_1}{D_1B}=1$. Now, again, this rewrites as $\frac{F_1R}{RE_1}\cdot\left(\frac{E_1D_1}{ED}: \frac{D_1F_1}{DF}\right)\cdot\frac{CD_1}{D_1B}=1$. Equivalently, (1) $\frac{F_1R}{RE_1}\cdot\left(\frac{E_1D_1}{ED}: \frac{F_1D_1}{FD}\right)\cdot\frac{CD_1}{D_1B}=1$. Now, since $D_1E_1\parallel DE$, Thales yields $\frac{E_1D_1}{ED}=\frac{CE_1}{E_1E}$. In other words, $\frac{E_1D_1}{ED}=\frac{CE_1}{E_1E}\cdot\frac{E_1E}{AE}\cdot\frac{AE}{CE}$. Another application of Thales using $D_1E_1\parallel DE$ shows that $\frac{CE_1}{E_1E}=\frac{CD_1}{D_1D}$, so this becomes $\frac{E_1D_1}{ED}=\frac{CD_1}{D_1D}\cdot\frac{E_1E}{AE}\cdot\frac{AE}{CE}$. Similarly, $\frac{F_1D_1}{FD}=\frac{BD_1}{D_1D}\cdot\frac{F_1F}{AF}\cdot\frac{AF}{BF}$. Thus, $\frac{E_1D_1}{ED}: \frac{F_1D_1}{FD}=\left(\frac{CD_1}{D_1D}\cdot\frac{E_1E}{AE}\cdot\frac{AE}{CE}\right): \left(\frac{BD_1}{D_1D}\cdot\frac{F_1F}{AF}\cdot\frac{AF}{BF}\right)$. Since EF || BC, by Thales we have $\frac{AE}{CE}=\frac{AF}{BF}$, so that this simplifies to $\frac{E_1D_1}{ED}: \frac{F_1D_1}{FD}=\left(\frac{CD_1}{D_1D}\cdot\frac{E_1E}{AE}\right): \left(\frac{BD_1}{D_1D}\cdot\frac{F_1F}{AF}\right)$ $=\frac{CD_1}{BD_1}\cdot\frac{E_1E}{AE}\cdot\frac{AF}{F_1F}=\left(-\frac{CD_1}{D_1B}\right)\cdot\left(-\frac{E_1E}{EA}\right)\cdot\left(-\frac{AF}{FF_1}\right)$ $=-\frac{CD_1}{D_1B}\cdot\frac{E_1E}{EA}\cdot\frac{AF}{FF_1}$. Hence, the equation (1) becomes $\frac{F_1R}{RE_1}\cdot\left(-\frac{CD_1}{D_1B}\cdot\frac{E_1E}{EA}\cdot\frac{AF}{FF_1}\right)\cdot\frac{CD_1}{D_1B}=1$. In other words, $-\left(\frac{F_1R}{RE_1}\cdot\frac{E_1E}{EA}\cdot\frac{AF}{FF_1}\right)\cdot\left(\frac{CD_1}{D_1B}\right)^2=1$. Thus, (2) we have $\frac{F_1R}{RE_1}\cdot\frac{E_1E}{EA}\cdot\frac{AF}{FF_1}=-1$ if and only if $\left(\frac{CD_1}{D_1B}\right)^2=1$. But according to the Menelaos theorem, applied to the triangle $AE_1F_1$ with the points R, E, F on its sidelines $E_1F_1$, $AE_1$ and $AF_1$, the equation $\frac{F_1R}{RE_1}\cdot\frac{E_1E}{EA}\cdot\frac{AF}{FF_1}=-1$ is equivalent to the assertion that the points R, E, F are collinear, i. e. that the point R lies on the line EF; since the point R is defined as the point of intersection of the lines $PD_1$ and $E_1F_1$, this assertion is thus equivalent to stating that the lines EF, $E_1F_1$ and $PD_1$ are concurrent. On the other hand, the equation $\left(\frac{CD_1}{D_1B}\right)^2=1$ is equivalent to $\frac{CD_1}{D_1B}=1$ (since $\frac{CD_1}{D_1B}=-1$ is impossible, unless the point $D_1$ is infinite, what we don't want to allow); and this is actually just the statement that the point $D_1$ is the midpoint of the segment BC. Hence, the result (2) rewrites as follows: The lines EF, $E_1F_1$ and $PD_1$ are concurrent if and only if the point $D_1$ is the midpoint of the segment BC. And the problem is solved... Darij

I do not see why D1 has to be the midpoint of BC. The configuration is projective. There should be no limitations on position of D1. The only changes needed is correcting typo to "line EF", and not "line E" and orientation of PBC as Darij did. ( Please, do not think that I translated from original Chineese text!) Thank you. M.T.

Here is my solution(?) : Using central projection transform triangles ABC and PBC into triangles with a point D into middle of BC. This will put point P onto A angle median.Then move line EF until it becomes midline, P will be in the A angle vertex. Lines E1F1 and PD1 transform into diagonals of an inscribed parallelogram. They intersect on the mid-line. Thank you. M.T.

armpist wrote: I do not see why D1 has to be the midpoint of BC. The configuration is projective. No, it isn't. There is a lot of parallels around in the problem statement. armpist wrote: Here is my solution(?) : Using central projection transform triangles ABC and PBC into triangles with a point D into middle of BC. This will put point P onto A angle median.Then move line EF until it becomes midline, P will be in the A angle vertex. Lines E1F1 and PD1 transform into diagonals of an inscribed parallelogram. They intersect on the mid-line. Not that I would understand it, but at least it is nice that you call it a "solution(?)", not a "solution". And as I said above, you can't just project everything expecting the parallel lines to remain parallel after the projection, and particularly you can't force the triangle DEF to be the medial triangle of the triangle ABC. Darij

Dear Darij, In your wonderful post http://www.mathlinks.ro/Forum/topic-59919.html you wrote about a line through midpoint: "..... It is not undefined. The midpoint between an Euclidean point and an infinite point is the latter infinite point (the Euclidean point is "insignificant"). Only the midpoint between two infinite points is undefined. And yes, assertion (c) very well holds with the point X being infinite. ...... " So it is possible to send BC to infinity (preserving parallels) and EF (chasing BC in parallel fashion) only half way there? Some experts have found it being hysterically funny. But really, where are all the lines that are parallel to the line being on it's way to infinity? Well, I think differently: the more unexplained phenomena - the better. People tend to like it because of the amusement and entertainment components of it... Thank you. M.T.

I have just read the official version of this problem. Here is my translation: Points D,E,F are on the sides BC,CA,AB of an acute-angled triangle ABC respectively, which satisfy EF//BC. $D_1$ is a point on side BC(different from B,D,C), through $D_1$ draw $D_1E_1//DE,D_1F_1//DF$, which intersect AC and AB at points $E_1,F_1$ respectively. Join $E_1F_1$. Construct,on the same side of A, triangle PBC, such that $\triangle PBC\sim\triangle DEF$. Join $PD_1$. Prove that $EF,E_1F_1,PD_1$ are concurrent. I know very well this must be the worse translation in the world...

mecrazywong wrote: I have just read the official version of this problem. Here is my translation: Points D,E,F are on the sides BC,CA,AB of an acute-angled triangle ABC respectively, which satisfy EF//BC. $D_1$ is a point on side BC(different from B,D,C), through $D_1$ draw $D_1E_1//DE,D_1F_1//DF$, which intersect AC and AB at points $E_1,F_1$ respectively. Join $E_1F_1$. Construct,on the same side of A, triangle PBC, such that $\triangle PBC\sim\triangle DEF$. Join $PD_1$. Prove that $EF,E_1F_1,PD_1$ are concurrent. That's funny, since this means the official version is wrong... darij

Hello guys, I think I see what is wrong. Darij put us in thinking that the triangle PBC can be constructed only by lines through B and C parallel to DE and DF. There is another triangle with vertex P still on the same line. And it has proper orientation. The problem is correct with only typo line E should be line EF. Now it all should work out, I just reduced the number of parallel lines for Darij. Thank you. M.T.

armpist wrote: Hello guys, I think I see what is wrong. Darij put us in thinking that the triangle PBC can be constructed only by lines through B and C parallel to DE and DF. There is another triangle with vertex P still on the same line. And it has proper orientation. The problem is correct with only typo line E should be line EF. You mean that triangle PBC should be similar to triangle DEF (with corresponding letters for corresponding vertices)? Try it, it won't hold. Darij

Dear Darij, PBC corresponds to FDE for direct similarity. I guess the problem statement just states similarity to make it harder to find the proper triangle. In my sketch it works out fine. The 3 lines intersect . Line EF if rather close to BC for angle D to be obtuse. If D and D1 on the same half of BC then PBC ~ EDF. Thank you. M.T.

A vectorial solution (orientated segments). The fixed points: $D$, $E$, $F$, $P\Longrightarrow$ the constant values $d$, $k$. The mobile points: $D_1$, $E_1$, $F_1$, $R\Longrightarrow$ the variable value $x$. $\frac{AE}{EC}=\frac{AF}{FB}=k\Longrightarrow \underline {\overline {\left| E=\frac{A+kC}{1+k},\ F=\frac{A+kB}{1+k}\right| }}.$ $\frac{BD}{DC}=d\Longrightarrow \underline {\overline {\left| D=\frac{B+dC}{1+d}\right| }}.$ $\frac{BD_1}{D_1C}=x\Longrightarrow \underline {\overline {\left| (1+x)D_1=B+xC\right| }}\ (1).$ $\triangle PBC\sim \triangle DFE\ (D\in AP)\Longrightarrow \frac{BP}{FD}=\frac{AB}{AF}=\frac{k+1}{k}\Longrightarrow$ $P=B+\frac{k+1}{k}(D-F)\Longrightarrow\underline {\overline {\left| k(1+d)P=-(1+d)A+(1+k)B+d(1+k)C\right| }}\ (2).$ $\frac{EE_1}{E_1C}=\frac{DD_1}{D_1C}=\frac{D_1-D}{C-D_1}=\frac{\frac{B+xC}{1+x} -\frac{B+dC}{1+d}}{C-\frac{B+xC}{1+x}}=\frac{x-d}{1+d}\Longrightarrow$ $E_1=\frac{E+\frac{x-d}{1+d} C}{1+\frac{x-d}{1+d}}=\frac{(1+d)E+(x-d)C}{1+x}\Longrightarrow$ $(1+x)E_1=(1+d)\frac{A+kC}{1+k}+(x-d)C\Longrightarrow$ $\underline {\overline {\left| (1+x)(1+k)E_1=(1+d)A+(k+x+kx-d)C\right| }}.$ $\frac{FF_1}{F_1B}=\frac{DD_1}{D_1B}=\frac{D_1-D}{B-D_1}=\frac{\frac{B+xC}{1+x}-\frac{B+dC}{1+d}}{B-\frac{B+xC}{1+x}}=\frac{d-x}{x(1+d)}\Longrightarrow$ $F_1=\frac{F+\frac{d-x}{x(1+d)} B}{1+\frac{d-x}{x(1+d)}}=\frac{x(1+d)\frac{A+kB}{1+k}+(d-x)B}{d(1+x)}\Longrightarrow$ $\underline {\overline {\left| d(1+k)(1+x)F_1=x(1+d)A+(kdx+kd+d-x)B\right| }}.$ $R\in EF\cap E_1F_1\Longleftrightarrow R=uE+(1-u)F=vE_1+(1-v)F_1\Longleftrightarrow$ $\frac{u}{kx+k+x-d}=\frac{1-u}{kdx+kd+d-x}=\frac{1}{k(1+d)(1+x)},$ $\frac v1=\frac{1-v}{d}=\frac{1}{1+d}\Longleftrightarrow$ $\underline {\overline {\left| (1+k)(1+d)(1+x)R=(1+d)(1+x)A+(kdx+kd+d-x)B+(kx+k+x-d)C\right| }}\ (3).$ ========================================================================== $D_1\in PR\Longleftrightarrow \left| \begin{array}{ccc} 0 & 1 & x \\ -(1+d) & 1+k & d(1+k) \\ (1+d)(1+x) & kdx+kd+d-x & kx+k+x-d \ \end {array}\right| =0$ $\Longleftrightarrow k(1+d)(x^2-1)=0\Longleftrightarrow x=1$, i.e. $D_1$ is the midpoint of the side $[BC]$. Remark. If the points $P$, $A$ are on the same side of the line $BC$ and $\triangle PBC\sim \triangle DEF$, ($PB\parallel DE$, $PC\parallel DF$) then the official version is correctly !

mecrazywong wrote: I have just read the official version of this problem. Here is my translation: Points $D$, $E$, $F$ are on the sides $BC$, $CA$, $AB$ of an acute-angled $\triangle{ABC}$ respectively, which satisfy $EF\parallel BC$. $D_1$ is a point on side $BC$(different from $B$, $D$, $C$), through $D_1$ draw $D_1E_1 \parallel DE, D_1F_1 \parallel DF$, which intersect $AC$ and $AB$ at points $E_1$, $F_1$ respectively. Join $E_1F_1$. Construct,on the same side of $A$, triangle $PBC$, such that $\triangle PBC\sim\triangle DEF$. Join $PD_1$. Prove that $EF,E_1F_1,PD_1$ are concurrent. Yes, this is the official version appeared in the published book by the Chinese coach teams.