# Geometry question -- several lines through parallelograms

## Homework Statement

On the picture, ##ABCD## is a parallelogram, ##(EF) // (AB) ##, and ##(GH) // (BC)##.

The problem is : show that lines ##(EB)##, ##(HD)##, and ##(IC)## either all meet in ##M##, or are parallel.

## The Attempt at a Solution

I've solved the problem introducing a coordinate system ##(D,\vec {DG},\vec {DA} )##, but this was the 'last chance' idea. I don't like it too much, it is a little messy, and I suspect there are simple ideas that could be used and that I didn't see. So my demand is rather for insights. How would you solve it ? Thanks !

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## Homework Statement

On the picture, ##ABCD## is a parallelogram, ##(EF) // (AB) ##, and ##(GH) // (BC)##.

The problem is : show that lines ##(EB)##, ##(HD)##, and ##(IC)## either all meet in ##M##, or are parallel.

## The Attempt at a Solution

I've solved the problem introducing a coordinate system ##(D,\vec {DG},\vec {DA} )##, but this was the 'last chance' idea. I don't like it too much, it is a little messy, and I suspect there are simple ideas that could be used and that I didn't see. So my demand is rather for insights. How would you solve it ? Thanks !
Parall AHGD less parall ABFE equals parall BCKJ. Three parallelograms similarly oriented in the plan; and adjoining neighbors. To find point J, just extend linear segment IC across segment AB. Side JK of the third parallelogram, is parallel to side BC. Any such a trio of parallelograms, induces a commun point of one of their respective diagonals. A theorem to be proven.

• geoffrey159
Hello. Thank you for your post.
What does parallelogram AHGD less parallelogram ABFE mean ? Can you draw a picture ?
What should be understood by orientation of several parallelograms ? I don't understand this.

To find point J, just extend linear segment IC across segment AB. Side JK of the third parallelogram, is parallel to side BC. Any such a trio of parallelograms, induces a commun point of one of their respective diagonals. A theorem to be proven.

It is possible after all that I understood nothing of what you said :-), but aren't you paraphrasing the problem statement ?

haruspex
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You can simplify the geometry by thinking about plane projections.

• theBin
What do you mean ? You may give your detailed solution as the exercise is already solved, it's just that I really feel that mine is longer than it needs to be.
It consists in fitting a coordinate system on the parallelogram, finding the cartesian equations of (EB), (HD), and (IC), and start working on the following alternative: two of these lines either meet at a point or are parallel. This alternative translates algebraically when one tries to solve the system of their cartesian equations.
If this algebraic condition is satisfied then the solution of the system is also a point of the third line, if not, they are mutually parallel.

haruspex
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What do you mean ? You may give your detailed solution as the exercise is already solved, it's just that I really feel that mine is longer than it needs to be.
It consists in fitting a coordinate system on the parallelogram, finding the cartesian equations of (EB), (HD), and (IC), and start working on the following alternative: two of these lines either meet at a point or are parallel. This alternative translates algebraically when one tries to solve the system of their cartesian equations.
If this algebraic condition is satisfied then the solution of the system is also a point of the third line, if not, they are mutually parallel.
If you take a photograph of a rectangle, but not square on, you get a parallelogram. You can reverse that by a suitable plane projection to convert a parallelogram to a rectangle. Straight lines are conserved, so without loss of generality you can consider only the case in which the shapes are rectangles.
However, I don't know whether that helps much. I suppose it would help with the algebraic approach.

• geoffrey159
You can simplify the geometry by thinking about plane projections.
I endeed simplified my sketches by switching to the Euclidean geometry, from the projective geometry. The transformation is legal, and would not affect the issue.

• geoffrey159
If you take a photograph of a rectangle, but not square on, you get a parallelogram. You can reverse that by a suitable plane projection to convert a parallelogram to a rectangle. Straight lines are conserved, so without loss of generality you can consider only the case in which the shapes are rectangles.
However, I don't know whether that helps much. I suppose it would help with the algebraic approach.

Intuititively, I understand the idea, but I don't immediately see how it simplifies the problem. In my mind, it amounts to replace the word 'parallelogram' by the word 'rectangle' in the problem statement.

Parall AHGD less parall ABFE equals parall BCKJ. Three parallelograms similarly oriented in the plan; and adjoining neighbors. To find point J, just extend linear segment IC across segment AB. Side JK of the third parallelogram, is parallel to side BC. Any such a trio of parallelograms, induces a commun point of one of their respective diagonals. A theorem to be proven.
erratum: "Parall AHGD less parall ABFE equals" is replaced, sorry for my mistake, by "We will construct" .
addendum: Let's switch to Euclidean geometry, in order to see rectangles; and let's remember the proven theorem that states that in any hexagon circumscribing/confining a circle, three diagonals are either concurrent or parallel. In our problem, in both projective and Euclidean geometries, two sides of any possible hexagon are always in a straight continuity (angle is 180 degrees). This criterion makes the "or are parallel" of the conjecture presented by Jeoffrey159, not receivable. If the audience accept the following simple proof based on the notion of order (term used in philosophy of mathematics), his conjecture becomes a new theorem. THE HARMONIC MEAN OF THE THREE DIAGONALS IS A CONSTANTE. A correlation to what happens in the above mentioned hexagon confining a circle, needs be done. The idea must be developped and the result could be translated in Bourbak's language & spirit, for purpose of publication.

@theBin : it is just a high school problem that I tried to brush up on my geometry, so it doesn't need advanced geometry to be solved.
However, you say that the 'or parallel' statement is wrong, but it's not what I find (by the means I used).

haruspex
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Intuititively, I understand the idea, but I don't immediately see how it simplifies the problem. In my mind, it amounts to replace the word 'parallelogram' by the word 'rectangle' in the problem statement.
Continuing the idea, we can normalise the outermost rectangle to a unit square. So now there is only pair of variables, the coordinates of the interior point. Certainly the algebra is now fairly straightforward. Similar triangles produce three equations relating this (x,y) to the coordinates of the intersection point, and it is easy to see that one can be deduced from the other two. I was hoping this would lead to some insight regarding the geometric argument, but I don't see it yet.

@theBin : it is just a high school problem that I tried to brush up on my geometry, so it doesn't need advanced geometry to be solved.
However, you say that the 'or parallel' statement is wrong, but it's not what I find (by the means I used).
By the written & graphic informations you presented, it is clear that your three diagonals have different orientations/angles. May I challenge you to present/show a case where three diagonals are parallel ?

erratum: "Parall AHGD less parall ABFE equals" is replaced, sorry for my mistake, by "We will construct" .
addendum: Let's switch to Euclidean geometry, in order to see rectangles; and let's remember the proven theorem that states that in any hexagon circumscribing/confining a circle, three diagonals are either concurrent or parallel. In our problem, in both projective and Euclidean geometries, two sides of any possible hexagon are always in a straight continuity (angle is 180 degrees). This criterion makes the "or are parallel" of the conjecture presented by Jeoffrey159, not receivable. If the audience accept the following simple proof based on the notion of order (term used in philosophy of mathematics), his conjecture becomes a new theorem. THE HARMONIC MEAN OF THE THREE DIAGONALS IS A CONSTANTE. A correlation to what happens in the above mentioned hexagon confining a circle, needs be done. The idea must be developped and the result could be translated in Bourbak's language & spirit, for purpose of publication.
Continuing the idea, we can normalise the outermost rectangle to a unit square. So now there is only pair of variables, the coordinates of the interior point. Certainly the algebra is now fairly straightforward. Similar triangles produce three equations relating this (x,y) to the coordinates of the intersection point, and it is easy to see that one can be deduced from the other two. I was hoping this would lead to some insight regarding the geometric argument, but I don't see it yet.
erratum: "Parall AHGD less parall ABFE equals" is replaced, sorry for my mistake, by "We will construct" .
addendum: Let's switch to Euclidean geometry, in order to see rectangles; and let's remember the proven theorem that states that in any hexagon circumscribing/confining a circle, three diagonals are either concurrent or parallel. In our problem, in both projective and Euclidean geometries, two sides of any possible hexagon are always in a straight continuity (angle is 180 degrees). This criterion makes the "or are parallel" of the conjecture presented by Jeoffrey159, not receivable. If the audience accept the following simple proof based on the notion of order (term used in philosophy of mathematics), his conjecture becomes a new theorem. THE HARMONIC MEAN OF THE THREE DIAGONALS IS A CONSTANTE. A correlation to what happens in the above mentioned hexagon confining a circle, needs be done. The idea must be developped and the result could be translated in Bourbak's language & spirit, for purpose of publication.
@theBin : it is just a high school problem that I tried to brush up on my geometry, so it doesn't need advanced geometry to be solved.
However, you say that the 'or parallel' statement is wrong, but it's not what I find (by the means I used).
Since there are more than one unknown, a complete solution would also contain other stuff, like an arithgmetic and a geometric means. I abandon this problem which is beyond my capacities and education.

haruspex
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By the written & graphic informations you presented, it is clear that your three diagonals have different orientations/angles. May I challenge you to present/show a case where three diagonals are parallel ?
That is not quite the point. There may be a not too difficult method which shows that either they intersect at a common point or are all three parallel. But it might require a more elaborate proof to rule out the parallel case.

• geoffrey159
haruspex
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Since there are more than one unknown, a complete solution would also contain other stuff, like an arithgmetic and a geometric means. I abandon this problem which is beyond my capacities and education.
As I wrote in post #11, if you accept the validity of my simplification-by-projection arguments, the algebra is not at all difficult. I'm happy to help you through it.
Labelling the vertices A, B, C left to right across the top, D, E, F across the middle, etc., ACIG forms a square. Make it a unit square with G as origin. Point E has coordinates (x,y), and the point of intersection of CD and BG is (u,v). Use similar triangles to express u and v in terms of x and y. Etc.

• geoffrey159
@theBin: Honestly you are attracting attention for nothing and I don't like it.
As I said, it is a high school problem, and anyone that knows two or three things about line intersections can get to the same result I got to, in minutes, following the process described in a previous post. So before the next angry post, I suggest you do it with simple methods and tell me weather it is possible or not.

berkeman
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