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Given $3$ different points on line $p$ : $A,B,C$, and $3$ different points on line $q$ : $A',B',C'$.

In addition: $A'B \parallel AB'$, and $A'C \parallel AC'$.

The lines $p$ and $q$ are intersecting at point $M$ which is different from the points that mentioned.

Need to prove that $BC' \parallel B'C$.

MY ATTEMPT: I tried to use triangles similarity to find another similarity for triangles $B'A'C$ and $ABC'$ but didn't succeeded.

Thank you.

may
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  • Please draw a picture and include it in your answer. – Babai Sep 28 '15 at 07:43
  • it's without a picture and i don't want to mislead someone.. – may Sep 28 '15 at 07:48
  • Pictures are misleading sometimes, but not most of the times. At least for these simple 3-dimensional geometric deductions one should draw pictures. See the answer posted has a figure to accompany with. – Babai Sep 28 '15 at 08:54
  • Related: http://math.stackexchange.com/questions/19692/euclidean-version-of-pappuss-theorem – Isomorphism Sep 28 '15 at 09:21

3 Answers3

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Consider the triangles $MA'B$ and $MB'A$, they are similar, so $$ \frac{MB'}{MA'}=\frac{MA}{MB}. $$ Consider now the triangles $MC'A$ and $MA'C$, they are similar, so $$ \frac{MC'}{MA'}=\frac{MA}{MC}. $$ Now, putting together these two equations we get $$ \frac{MC'}{MB'}=\frac{MB}{MC} $$ this means that the triangles $MB'C$ and $MC'B$ are similar, therefore $B'C||C'B$.

mrprottolo
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Pappus' theorem says, as illustrated below,

enter image description here

that let $3$ points $A$, $B$, $C$ on the straight line $p$ and $3$ points $A'$,$B'$, $C'$ on the straight line $q$ be given then if one constructs the straight lines joining $AB'$ and $BA'$ and their intersection point (fat point), and then the straight lines $BC'$ and $B'C$ and their intersection point (fat point) and then the straight lines $AC'$ and $CA'$ and their intersection point (fat point) then the said fat points points will lie on one straight line.

If two of the pairs of the straights lines constructed above are parallel then the third pair will be parallel too. This is because in this case the first two crossing points will be on the infinitely remote straight line, that is the crossing point of the third pair will have to be on that same line, which means parallelism as shown in the following figure:

enter image description here

This is the configuration desciben in the OP.

zoli
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Suggesting a 3D solution where p and q are parallel generators of a cylinder meeting at infinity,

All are parallelograms as per labels. Fold the paper along generator CB C'B' so that bottom line new C'B' ( labels retained) falls on p making a 3D cylinder. B'C and BC' are on the same line.Parallelism holds for helices.

Works with 0,1,2 multiple copy/fold.

enter image description here

Narasimham
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