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Similarity between triangles

by disregardthat
Tags: similarity, triangles
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disregardthat
#1
Jan19-08, 04:32 PM
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Hi, do anyone know a proof of this converse:

"If, A,B,C,D,E and F are points in the plane and [tex]\frac{AB}{BC}=\frac{DE}{EF}[/tex], then triangles ABC and DEF are similar."
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EnumaElish
#2
Jan19-08, 04:50 PM
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What is your definition of similar?
disregardthat
#3
Jan19-08, 04:54 PM
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Oh, with similiar I mean that the triangles are equiangular.

With [tex]\angle ABC = \angle DEF[/tex] and [tex]\angle ABC = \angle DEF[/tex]

EnumaElish
#4
Jan19-08, 05:02 PM
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Similarity between triangles

Quote Quote by Jarle View Post
Oh, with similiar I mean that the triangles are equiangular.

With [tex]\angle ABC = \angle DEF[/tex] and [tex]\angle ABC = \angle DEF[/tex]
Do you mean /_ABC = /_ DEF and /_BCA = /_EFD ? (You repeated yourself.)
disregardthat
#5
Jan19-08, 05:55 PM
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Yes, I meant the last thing you said. (Getting late.)
jdg812
#6
Jan19-08, 06:11 PM
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Quote Quote by Jarle View Post
Hi, do anyone know a proof of this converse:

"If, A,B,C,D,E and F are points in the plane and [tex]\frac{AB}{BC}=\frac{DE}{EF}[/tex], then triangles ABC and DEF are similar."
No one knows a proof, because such a proof doesn’t exist.

PS
But in the case AB:BC:CA = DE:EF:FD, proof is trivial
disregardthat
#7
Jan19-08, 06:21 PM
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Forgot to mention that they share the angle between the sides AB and BC, and DE and EF.

Oh, and the expression should not look like that either...

It's [tex]\frac{AB}{DE}=\frac{EF}{BC}[/tex]

EDIT: Anyway, I have proven it now...
jdg812
#8
Jan19-08, 06:29 PM
P: 91
Quote Quote by Jarle View Post
Forgot to mention that they share an angle.
Still not enough for similarity...

If you forget as well to mention that equal angles are /_ABC and /_ DEF, proof is trivial.
If other angles, proof doesn't exist.
disregardthat
#9
Jan19-08, 06:33 PM
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There has been to many faults here. I will sum it up and see what I find out!
jdg812
#10
Jan19-08, 06:35 PM
P: 91
Quote Quote by Jarle View Post
Forgot to mention that they share the angle between the sides AB and BC, and DE and EF.
Oh, and the expression should not look like that either...

It's [tex]\frac{AB}{DE}=\frac{EF}{BC}[/tex]

EDIT: Anyway, I have proven it now...
After your editions (bolded) proof is not possible

It must be

[tex]\frac{AB}{DE}=\frac{BC}{EF}[/tex]

OR

[tex]\frac{AB}{BC}=\frac{DE}{EF}[/tex]
disregardthat
#11
Jan19-08, 07:08 PM
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Ok, I have summed it up now. I am trying to prove the converse of the intersecting chords theorem, in the case where the point of intersection is inside the circle...

We have that the line segments AB and CD meet at X. So the opposite angles at X are equal. We translate the triangles to the triangle with sides XAD, XCB and with AB and DC joined. Now the angle at X is a, and [itex]\frac{XA}{XC}=\frac{XD}{XB}[/itex]. This is a sufficient condition for the triangles XAC and XDB to be similar, as corresponding sides have the same ratio, and the included angle is equal. Thus is the angle XDB equal to the angle XAC, and by the converse of the angles subtended by the same arc theorem, ACBD are concyclic points, so AB and CD are chords in a circle. Does this look ok to you? The other cases of the converse goes something in the same way.


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