Prove that the ratio of the length of a diagonal to that of its corresponding side

[.d9n.]

Pentagon diagonal - parallel ratio proof

Homework Statement

Suppose P is a convex pentagon such that each diagonal is parallel to one side. Prove that the ratio of the length of a diagonal to that of its corresponding side is the same for all five diagonals, and compute this ratio

Homework Equations

The Attempt at a Solution

I think they mean a pentagon where each vertices is connected by an edge. So there is a pentagon in the middle surrounded by obtuse and acute triangles. So if you label the sides of the original pentagon each side is say a+b, with the bases of the acute triangles are say b and the sides are say a. Then the ratio of the paralleled sides is a+b:2a+b? Not really sure how i am meant to prove this though?

 

[Mark44]

Homework Statement

Suppose P is a convex pentagon such that each diagonal is parallel to one side. Prove that the ratio of the length of a diagonal to that of its corresponding side is the same for all five diagonals, and compute this ratio

Homework Equations

The Attempt at a Solution

I think they mean a pentagon where each vertices is connected by an edge.

No, I don't think so. Each pair of adjacent vertices is connected by an edge, but nonadjacent vertices are not connected.

So there is a pentagon in the middle surrounded by obtuse and acute triangles.

No. It's just a pentagon, a five-sided figure. It's a convex pentagon, which means that the interior angles are all less than 180°.
The pentagon is not surrounded by triangles of any kind. I don't know where you got that idea.

So if you label the sides of the original pentagon each side is say a+b, with the bases of the acute triangles are say b and the sides are say a. Then the ratio of the paralleled sides is a+b:2a+b? Not really sure how i am meant to prove this though?

KEEP IT SIMPLE. Just draw a pentagon, and draw some of the diagonals, keeping in mind what it says about them being parallel to another side of the pentagon.

 

[.d9n.]

Oh I meant the triangles are in the interior, when you draw the diagonals. So if a pentagons vertices are labelled 1 to 5, then say the length of 3-4 is a+b, then the parallel diagonal is length 2a+b?

 

[checkitagain]

Suppose P is a convex pentagon such that each diagonal is
parallel to one side. Prove that the ratio of the length of a
diagonal to that of its corresponding side is the same for
all five diagonals, and compute this ratio

.d9n.,

would it be the case that this convex pentagon would have to
be equilateral?


Furthermore, would it be the case that this convex pentagon
would have to be a regular pentagon?

 

[.d9n.]

im not sure this is all the info i have

 

[Mark44]

Oh I meant the triangles are in the interior, when you draw the diagonals. So if a pentagons vertices are labelled 1 to 5, then say the length of 3-4 is a+b, then the parallel diagonal is length 2a+b?

Why are you setting the length of the edge between vertice 3 and 4 to a + b? The parallel diagonal would be between vertices 2 and 5. Why would you think this would be 2a + b?

What you are doing is very confusing.

 

[.d9n.]

yer i know, this is where i got the triangles and ratios from

http://www.jimloy.com/geometry/pentagon.htm

 

[Mark44]

The figure in the page in the link appears to be an equilateral pentagon, which is information you are not given. From the information you are given, you cannot assume that all the sides are equal. It may be that they are, but you would need to show that, using geometry and trig.

 

[.d9n.]

oh right, any ideas then on how i do that, I am a bit lost

 

[checkitagain]

The figure in the page in the link appears
to be an equilateral pentagon,...

Not only does the pentagon in the link appear to be equilateral, it also
appears to be equiangular.

 

[.d9n.]

so any ideas on how to go about proving this? i have no idea
thanks

 

[Mark44]

I would start by drawing a pentagon, and a couple of its diagonals. Use the given information that each diagonal is parallel to one side. This means that a given side of the pentagon, its two adjacent sides, and the diagonal, form a trapezoid. See where that takes you.

 

[.d9n.]

ok so if we label each trapezoid:

if we assign lengths to each edge

ab=v, bc=w, cd=x, de=y, ea=z,

now the diagonals

ac=f, ad=g, be=h, bd=i, ce=j

1. abcd ratio = bc/ad = w/g
2. bcde ratio = cd/be = x/h
3. abde ratio = ae/bd = z/i
4. acde ratio = de/ac = y/f
5. abce ratio = ab/ce = v/j

not sure how to prove all the ratios are the same without it being equilateral

 

[Mark44]

You need to use the given information that each diagonal is parallel to one of the sides. From that, you need to establish relationships between the angles formed by the interior triangles.

 

[.d9n.]

do u mean by putting diagonals in the trapezoid?
unsure how to do this if i can't assume the the shape is equilateral, as surely the angles could be anything depending on the shape of the original pentagon.

 

[.d9n.]

if your make the middle of the pentagon 'o' say then, its 360n degrees around it divide by 5 leaves each angle aob, boc, cod, doe, eoa is 72 degrees?
making the angles abc bcd cde dea eab all 108? don't know if this is of any help though?

 

[.d9n.]

which potentially means angle ead and similar are 36 degrees?

 

[Mark44]

if your make the middle of the pentagon 'o' say then, its 360n degrees around it divide by 5 leaves each angle aob, boc, cod, doe, eoa is 72 degrees?

Only if it's a regular pentagon, which is information you're not given.

making the angles abc bcd cde dea eab all 108? don't know if this is of any help though?

 

[.d9n.]

that's what i thought, so I am unsure of how to work out the angles if we don't know what type of pentagon it is

 

[.d9n.]

surely the angles will change depending on what type of pentagon it is, i.e. isosceles, equilateral, or neither. I've looked all over trying to find a way to do it, yet no luck

 

[SammyS]

You can't assume that the pentagon is equilateral. It doesn't have to be equilateral for this theorem to be true.

You can't pick some arbitrary convex pentagon either. It's likely that such a pentagon would not fulfill the condition that each diagonal is parallel to one side.

This looks to me to be very difficult to prove -- unless there is some nice trick.

 

[.d9n.]

can i ask what theorem you are referring to?

Any hints on where to look for this 'nice trick', is it enclosed in the theorem you refer to?

 

[.d9n.]

so when a line intersects the two parallel lines, where the top line is intersected the the interior angle is say 'a' so the interior angle on the bottom line is 180-a?

 

[.d9n.]

but then i still don't know how that helps me as i can't assume that another intersecting line intersects the parallel line at the same angle. I am stuck

 

[SammyS]

can i ask what theorem you are referring to?

Any hints on where to look for this 'nice trick', is it enclosed in the theorem you refer to?

I was referring to the conjecture you're trying to prove here.

I have no idea about where to look for such a trick.

 

[.d9n.]

oh right, so any idea about what i should do next