Impossible Triangle?

  • #1
My Latin teacher is a retired nuclear physicist. He is a very smart guy. Sometimes, he shows us things called ABL (anything but latin) like math or logic. Its a fun class. Anyway, there is one geometry problem he gave us that has been bugging me.

Look at this diagram on this link.

Prove That The Triangle is Isoceles. Be careful no to assume things.

Answers and Replies

  • #2
This is a notorious question. It can be done be angle chasing and repeated application of the sin rule but you don't want to do that. There is also a clever trick, but I don't remember it off the top of my head. Try searching for 'isocoles triangle martin gardner' as it's one of his favourite puzzles and someone may have put the solution on the web.

Actually I'm not convinced it's the same problem bit it's the same diagram
  • #3
Okay thanks.
  • #4
I'm not sure if math proofs accept logic yet, but I would just go about it like:

It must be isosceles because if CB grew or contracted (without CA changing equally), DB would no longer bisect FBA, and if CA grew or contracted (without CB changing equally), FA would no longer bisect DAB.

I hope you see what I mean, I tend to think of lines moving like an animation.
  • #5
I know exactly what you mean. But you have to use a geometric proof. Prove it!
  • #6
I can't see the image.
  • #7
Originally posted by Decker
It must be isosceles because if CB grew or contracted (without CA changing equally), DB would no longer bisect FBA, and if CA grew or contracted (without CB changing equally), FA would no longer bisect DAB.
The logic is correct, but I think it's the other way around. If CB changed, FA wouldn't bisect DAB and if CA changed, DB wouldn't bisect FBA.
  • #8
Ok, how about this?

Let [tex]\alpha[/tex] be FAB;
Let [tex]\beta[/tex] be DBA;

In the trapeze ADFB we draw heights from both D and F. These two lines (let's call them DG and FK) must be of equal length:

[tex]DG = FK[/tex]

[tex]BD\sin \beta = AF\sin \alpha[/tex]
[tex]BD = AF[/tex] (given)

[tex]\sin \beta = \sin \alpha[/tex]
[tex]\beta = \alpha[/tex] (both are smaller than 90 degrees)
[tex]2\beta = 2\alpha[/tex]


Edit: Ok, maybe I forgot to prove that ADFB is a trapeze.
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  • #9
Chen, your first statement isn't correct. You can't make that first statement equal because you are dealing with two different triangles. To make that equal is to assume they are equal, and therefore it is no surprise that you determine that alpha = beta in the end. You assumed it at the beginning.

I have worked out how to prove this. It is by no means easy. Here it is:

The bisectors of a triangle's angles intercept at the 'incentre' of the triangle, the centre of the inscribed circle. In this case, E is the incentre, and CE bisects angle C.

Now, project CE onto AB and call that point of intersection G. Now, the inscribed circle passes through points D, F and G. Therefore, since DE, EF and EG are thus radii of that inscribed circle, they are equal to each other.

Now, since it was given that AF=BD, and it has been proven that DE=EF, we can thus reason that EB=EA.

Now, in triangles DEA and FEB:
1: DE = EF (proven)
2: AE = EB (proven)
3: angle DEA = angle BEF (vertically opposite)
Therefore triangles DEA and FEB are congruent due to having two sides and the included angle of equal size/length, and therefore angle DAE = angle FBE.

It can then be seen that CAB = CBA, and the triangle is isosceles.

-edit-: Sorry, my bad. I jumped the gun majorly there. The inscribed circle does not go through D, F and G at all. After saying that, I then deduced that DE = EF, which was wrong. Sorry about that.
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  • #10
Originally posted by vertigo
Chen, your first statement isn't correct.
I know, I realized it after finishing everything. :smile:
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  • #11
I read somwhere that you have to use the Law of Sines to prove the triangle. If you can prove that angle CDB is congruent to angle CFA, then you can prove the big triangle is congruent.

We can do this by proving that triangle CDB is congruent to triangle CFA. Angle C is congruent to angle C relfexive property. And it is given that DB is congruent to AF. We have to prove one more line segment or another angle.

Try using the LAw of Sines. I don't know where to use it, but I heard a rumor that you can use a geometric proof with the LAW of Sines.
  • #12
here's a purported proof. I've not checked it

==> geometry/bisector.p <==
If two angle bisectors of a triangle are equal, then the triangle is
isosceles (more specifically, the sides opposite to the two angles
being bisected are equal).

==> geometry/bisector.s <==
The following proof is probably from Altshiller-Court's College
Geometry, since that's where I first saw the problem.

Let the triangle be ABC, with angle bisectors BE and CD.
Let F be such that BEFD is a parallelagram.
Let x = measure of angle CBE = angle DBE,
y = measure of angle BCD = angle DCE,
x' = measure of angle EFC,
y' = measure of angle ECF.
(You will probably want to draw a picture.)

Suppose x > y. Consider the triangles EBC and DCB. Since BC = BC and
BE = CD, we must have CE > BD. Now, since BD = EF, we have that CE >
EF, so that x' > y'. Thus x+x' > y+y'. But, triangle FDC is
isosceles, since DF = BE = DC, so x+x' = y+y', a contradiction.
Similarly, we cannot have x < y. Therefore the base angles of ABC are
equal, making ABC an isosceles triangle. QED
  • #13
Sounds about right... long, a bit messy even, but right. :)

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