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Look at this diagram on this link.

http://www.geocities.com/threewood14/

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

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- #1

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Look at this diagram on this link.

http://www.geocities.com/threewood14/

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

- #2

matt grime

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Actually I'm not convinced it's the same problem bit it's the same diagram

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Okay thanks.

- #4

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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.

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I know exactly what you mean. But you have to use a geometric proof. Prove it!

- #6

NateTG

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I can't see the image.

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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.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.

- #8

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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]

Right?

*Edit*: Ok, maybe I forgot to prove that ADFB is a trapeze.

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]

Right?

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- #9

verty

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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 suprise 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.

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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I know, I realized it after finishing everything.Originally posted by vertigo

Chen, your first statement isn't correct.

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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

matt grime

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==> 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

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Sounds about right... long, a bit messy even, but right. :)

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