A Simple Math Problem. (Relatively Speaking)

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A Simple Math Problem. (Relatively Speaking)
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 as well as Apotheosis.
Look at this diagram on this link.
http://www.geocities.com/threewood14/
Prove That The Triangle is Isoceles. Be careful no to assume things as we have.
P.S. I know that this is philosophy, but this problem has been bugging us and I know that there are a lot of smart people here. I just could not resist!
Look at this diagram on this link.
http://www.geocities.com/threewood14/
Prove That The Triangle is Isoceles. Be careful no to assume things as we have.
P.S. I know that this is philosophy, but this problem has been bugging us and I know that there are a lot of smart people here. I just could not resist!
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 Jung He Fah Toy
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Doesn't this come down to prooving either Pythagoras or that a triangle with two equal angles has two equal sides? I mean I would do that for you but its not really the proof that matters of course.
Also we can say that E1 = E3 and E2=E4, also A1=A2=B1=B2, so triangle ADE = BFE and AD=BF. D1 is F1 so D2 = F2 and CQD = CQF so CD = CF and both sides are equal in length?
Also we can say that E1 = E3 and E2=E4, also A1=A2=B1=B2, so triangle ADE = BFE and AD=BF. D1 is F1 so D2 = F2 and CQD = CQF so CD = CF and both sides are equal in length?
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Here is my solution  but note that I am using ^ to indicate angles (as opposed to triangles).
To prove that ABC is isoceles, we need to prove at least one of the following:
AC = BC
^CAB = ^CBA
So 
Let us add to the diagram segment DF.
This gives us quadrilateral ADFB.
Since DB and AF are congruent diagonals of ADFB which bisect the base angles DAB and FBA,
ADFB must be an isoceles trapezoid.
The lower base angles of any isoceles trapezoid are equal, therefore
^DAB = ^FBA
Since
^DAB = ^CAB and ^FBA = ^CBA
it goes to show that
^CAB = ^CBA
Thus, ABC must be isoceles.
To prove that ABC is isoceles, we need to prove at least one of the following:
AC = BC
^CAB = ^CBA
So 
Let us add to the diagram segment DF.
This gives us quadrilateral ADFB.
Since DB and AF are congruent diagonals of ADFB which bisect the base angles DAB and FBA,
ADFB must be an isoceles trapezoid.
The lower base angles of any isoceles trapezoid are equal, therefore
^DAB = ^FBA
Since
^DAB = ^CAB and ^FBA = ^CBA
it goes to show that
^CAB = ^CBA
Thus, ABC must be isoceles.
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But how do you know that DF is parallel to the base segment. If it were, then angle CAB must be congruent to angle CBA. You cannot assume this to prove it. In other words, you are assming that the two base angles are congruent. This is what we are trying to prove!
To show you what I mean,
I heard a rumor that you have to use the Law of Sines to prove it.
SinA/a = SinB/b = SinC/c
is the law of sines
To show you what I mean,
You are assuming that the angles bisected are congruent. Their bisectors may be equal in length, but the angles may not be the same. Therefore the top of the quadrilateral is not paralell to AB.Since DB and AF are congruent diagonals of ADFB which bisect the base angles DAB and FBA,
ADFB must be an isoceles trapezoid.
I heard a rumor that you have to use the Law of Sines to prove it.
SinA/a = SinB/b = SinC/c
is the law of sines
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An isoceles trapezoid has the following properties:
it is a quadrilateral
it has congruent diagonals
its diagonals bisect its internal angles
the lower base angles are congruent
the upper base angles are congruent
the pair of nonparallel edges are congruent
the pair of parallel edges are not congruent
Thus, if ADFB can be proven to be an isoceles trapezoid, ADFB has all these properties. That would show that DF is parallel to the base segment, that the base angles are equal, and that my solution holds.
From the information given, we know that ADFB has congruent diagonals which bisect the lower base angles. I have taken this to prove that ADFB is an isoceles trapezoid; you are questioning whether that is proof enough.
In other words, you are asking:
Can't there exist a quad which has congruent diagonals that bisect the internal angles, but which is NOT an isoceles quadrilateral?
I've tried looking that up, but so far haven't managed to find anything about it. From testing it so far, it seems the answer to the question is no  but if anyone can show otherwise, I'm open to persuasion.
...
So  I will accept, for the moment, that may proof might not hold (though it isn't ready to be tossed out the window yet!).
Keep in mind that there may be more than one solution.
Meanwhile, I'll see what I can do with the law of sines...
it is a quadrilateral
it has congruent diagonals
its diagonals bisect its internal angles
the lower base angles are congruent
the upper base angles are congruent
the pair of nonparallel edges are congruent
the pair of parallel edges are not congruent
Thus, if ADFB can be proven to be an isoceles trapezoid, ADFB has all these properties. That would show that DF is parallel to the base segment, that the base angles are equal, and that my solution holds.
From the information given, we know that ADFB has congruent diagonals which bisect the lower base angles. I have taken this to prove that ADFB is an isoceles trapezoid; you are questioning whether that is proof enough.
In other words, you are asking:
Can't there exist a quad which has congruent diagonals that bisect the internal angles, but which is NOT an isoceles quadrilateral?
I've tried looking that up, but so far haven't managed to find anything about it. From testing it so far, it seems the answer to the question is no  but if anyone can show otherwise, I'm open to persuasion.
...
So  I will accept, for the moment, that may proof might not hold (though it isn't ready to be tossed out the window yet!).
Keep in mind that there may be more than one solution.
Meanwhile, I'll see what I can do with the law of sines...
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We know that the two diagonals are congruent. But how do you prove that the angles that they bisect are congruent? Its the whole point of the problem. Like I said in my 1st post, we cannot assume anything.
I tried almost the exact proof and showed it to my teacher. That is what he pointed out.
P.S. Just to let you guys know, I'm trying to figure this out just so my teacher will give me a buck. LOL! Its a challenge though, huh?
I tried almost the exact proof and showed it to my teacher. That is what he pointed out.
P.S. Just to let you guys know, I'm trying to figure this out just so my teacher will give me a buck. LOL! Its a challenge though, huh?
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 Lex
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This is only the case if the bisected angles are congruent. This is only the case if the triangle ABC is isosceles, which is what is to be proven. Threewood is correct; you are assuming what is to be proven, without realizing it.Raya wrote:From the information given, we know that ADFB has congruent diagonals which bisect the lower base angles. I have taken this to prove that ADFB is an isoceles trapezoid
I, Lex Llama, super genius, will one day rule this planet! And then you'll rue the day you messed with me, you damned dirty apes!
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Another hint: You also know that AB is (trivially) congruent to itself. So, you know that the triangles ABD and BAF have two congruent sides; AB with AB, and DB with AF.threewood14 wrote:We know that the two diagonals are congruent.
Connect this with my previous hint, and you're halfway there.
I, Lex Llama, super genius, will one day rule this planet! And then you'll rue the day you messed with me, you damned dirty apes!

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Here is what I am working on.
Look at triangles ACF and BCD. If we can prove that these triangles are congruent, then we prove that triangle ABC is isosceles! In order to prove that a triangle is congruent to another, we must prove a combination of either 3 angles or sides. For example we can have ASA or SAS. Let's try to do this for these 2 triangles.
First, both triangles share a common angle. This angle is angle ACB. Angle ACB is congruent to angle ACB by the reflexive property. We know that segments AF and BD are congruent. This information is given. In order to prove that the 2 triangles are congruent, we must prove one of the following:
CD is congruent to CF
AC is congruent to CB
Angle CAF is congruent to angle CBD
Angle CDB is congruent to angle CFA
If anyone of these can be proven, then all the others can be proven true by CPCTC (Congruent Parts of Congruent Triangles are Congruent)
I'm not sure if the Law of Sines works because I haven't figured it out for myself. It's tougher than one thinks...
Look at triangles ACF and BCD. If we can prove that these triangles are congruent, then we prove that triangle ABC is isosceles! In order to prove that a triangle is congruent to another, we must prove a combination of either 3 angles or sides. For example we can have ASA or SAS. Let's try to do this for these 2 triangles.
First, both triangles share a common angle. This angle is angle ACB. Angle ACB is congruent to angle ACB by the reflexive property. We know that segments AF and BD are congruent. This information is given. In order to prove that the 2 triangles are congruent, we must prove one of the following:
CD is congruent to CF
AC is congruent to CB
Angle CAF is congruent to angle CBD
Angle CDB is congruent to angle CFA
If anyone of these can be proven, then all the others can be proven true by CPCTC (Congruent Parts of Congruent Triangles are Congruent)
I'm not sure if the Law of Sines works because I haven't figured it out for myself. It's tougher than one thinks...
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 Lex
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You're focusing on the wrong triangles. Focus on ABD and BAF; they have a side in common. Connect this with my other hints. You don't need the Law of Sines. Trust me.threewood14 wrote:Here is what I am working on.
Look at triangles ACF and BCD. If we can prove that these triangles are congruent, then we prove that triangle ABC is isosceles!
I, Lex Llama, super genius, will one day rule this planet! And then you'll rue the day you messed with me, you damned dirty apes!

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I do not think that triangles with two common sides must themselves be congruent. You also need an angle or another side. I can have two triangles. 4 of the 6 sides are congruent. Okay its hopeless trying to explain in words. Click here to see another diagram...triangles with two congruent sides must themselves be congruent
www.geocities.com/threewood14
It is harder than one thinks...
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 Lex
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Oops... you're right! My bad!threewood14 wrote:I do not think that triangles with two common sides must themselves be congruent. You also need an angle or another side.triangles with two congruent sides must themselves be congruent
Hmmm, back to the drawing board.
I, Lex Llama, super genius, will one day rule this planet! And then you'll rue the day you messed with me, you damned dirty apes!
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 Apotheosis
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Ok, here are our options and/or possible scenarios that may or may not be true:
1) It is a genuine case of Farmer Brown. (i.e.  the triangle isn't even a triangle at all)
2) We do not have enough information.
3) We are unaware of the methods that are required to prove that this triangle is isosceles.
4) We cannot use a geometric proof to prove that this triangle is isosceles as we have been trying to do.
5) We cannot prove the triangle to be isosceles, period. (According to Mr. Sytek  the man who gave us this problem  we will never be able to figure it out.)
6) We can devote our entire lives to solving this problem which may not even be solvable, thus all of our hard efforts would be in vein if that were the case.
Well there you have it! Take your pick folks! What'll it be?
1) It is a genuine case of Farmer Brown. (i.e.  the triangle isn't even a triangle at all)
2) We do not have enough information.
3) We are unaware of the methods that are required to prove that this triangle is isosceles.
4) We cannot use a geometric proof to prove that this triangle is isosceles as we have been trying to do.
5) We cannot prove the triangle to be isosceles, period. (According to Mr. Sytek  the man who gave us this problem  we will never be able to figure it out.)
6) We can devote our entire lives to solving this problem which may not even be solvable, thus all of our hard efforts would be in vein if that were the case.
Well there you have it! Take your pick folks! What'll it be?
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 Lex
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I tried several ways, to no avail. I suspect the constraints are insufficient to force the triangle to be isosceles. If this is the case, then it should be possible to find a counterexample that meets all the constraints and yet is not isosceles. This would at least prove that it's a trick question.Jung He Fah Toy wrote:If either of the triangles Lex and threewood were pointing out can be proven congruent, then the problem can be solved!
I, Lex Llama, super genius, will one day rule this planet! And then you'll rue the day you messed with me, you damned dirty apes!

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 Apotheosis
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Actually, wrote that question on my drill today before I handed it in. I said: "Can you use the law of sines to prove this triangle to be isosceles? Can you use a geometric proof to prove it is isosceles? Is it even possible?"
I hope he gets back to me tomorrow, if not we will aks him personally.
I hope he gets back to me tomorrow, if not we will aks him personally.
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 Jung He Fah Toy
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I might have something, but I'm not sure how valid it is.
First, let's start with a slightly different problem. Let's assume that DB and AF are "altitudes" of the triangle. That is, DB meets AC at a 90 degree angle, and AF meets BC at a 90 degree angle. It can be shown that if DB and AF are congruent, then the triangle must be isosceles. (I don't know the proof for this; I found it in a geometry book, and the proof wasn't included.)
OK, if you have that isosceles triangle with congruent altitudes DB and AF, then by pivoting DB and AF by the same number of degrees (and extending the line segments as necessary), you should be able to bisect the angles ABC and BAC. DB and AF will remain congruent. This proves that if ABC is an isosceles triangle with two line segments bisecting the congruent angles, then the bisectors will be congruent as well.
Now, can the converse be proven by converting the bisectors to altitudes? If so, the problem is solved. I'll leave this as an exercise for the reader.
First, let's start with a slightly different problem. Let's assume that DB and AF are "altitudes" of the triangle. That is, DB meets AC at a 90 degree angle, and AF meets BC at a 90 degree angle. It can be shown that if DB and AF are congruent, then the triangle must be isosceles. (I don't know the proof for this; I found it in a geometry book, and the proof wasn't included.)
OK, if you have that isosceles triangle with congruent altitudes DB and AF, then by pivoting DB and AF by the same number of degrees (and extending the line segments as necessary), you should be able to bisect the angles ABC and BAC. DB and AF will remain congruent. This proves that if ABC is an isosceles triangle with two line segments bisecting the congruent angles, then the bisectors will be congruent as well.
Now, can the converse be proven by converting the bisectors to altitudes? If so, the problem is solved. I'll leave this as an exercise for the reader.
I, Lex Llama, super genius, will one day rule this planet! And then you'll rue the day you messed with me, you damned dirty apes!

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I almost got it. Sin A/AF = Sin B/DB. AF is con. to DB. So SinA/AF = SinB/AF. THerefore angles CAB and CBA are congruent. Its not correct though trust me.
I found online that you have to use the apex, which in this case is angle C, and the law of sines to prove this. Look at triangles CDB and CFA. Try to do something with this so prove that these triangles are congruent.
I found online that you have to use the apex, which in this case is angle C, and the law of sines to prove this. Look at triangles CDB and CFA. Try to do something with this so prove that these triangles are congruent.
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 Apotheosis
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Kyle, I have something that will bust this mystery wide open (I hope). I showed this problem to my parents, and my mom went and got her old high school math book. Sure enough, the problem that Mr. Sytek gave us is in there! However, in the book, it is given that ?ABC is isosceles and you must prove AF and DB to be congruent. Now, the problem we are trying to solve is the exact opposite of that! We are trying to prove ?ABC to be isosceles given that AF and DB are congruent. The way I see it, if we can somehow "reverse engineer" the proof in the book, then we can solve this problem once and for all. I will bring the book into class tomorrow so you can see it. Hopefully we will be able to figure it out! Hehehehe....
P.S.  Mr. Sytek simply wrote on my paper: It is possible, but it is very difficult to prove it.
P.S.  Mr. Sytek simply wrote on my paper: It is possible, but it is very difficult to prove it.
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Here is a geometric one, but it is indirect; it uses a reductio ad absurdum argument.threewood14 wrote:I'm glad you found that, but we are trying to prove it with a geometric proof...Sorry
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Yes but we already know this. We know that the triangle can exist no other way.
We are trying to prove that it is an isosceles triangle not by proving that if it were any other way it would not be, but by using the info provided. In other words, we need a
this is this
threrfore this is this
therefore this equals this
therefore this is this
and so on
The challenge is proving it this way. It is possible with the law of sines. Our Latin teacher told us (who by the way is a retired nuclear physicist) that is is possible using the law of sines. It is harder than one thinks...
We are trying to prove that it is an isosceles triangle not by proving that if it were any other way it would not be, but by using the info provided. In other words, we need a
this is this
threrfore this is this
therefore this equals this
therefore this is this
and so on
The challenge is proving it this way. It is possible with the law of sines. Our Latin teacher told us (who by the way is a retired nuclear physicist) that is is possible using the law of sines. It is harder than one thinks...
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 Apotheosis
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Kyle...the answer is the SteinerLehmus theorem. Look:
Given:
AF = DB
AF bisects <CAB
DB bisects <CBA
Step 1:
AF = DB
Reason:
Given
Step 2:
Segment AF bisects angle CAB and extends to the side opposite it (Side CB)
Reason:
Given in diagram
Step 3:
Segment DB bisects angle CBA and extends to the side opposite it (Side CA)
Reason:
Given in diagram
Step 4:
Triangle ABC is isosceles.
Reason:
SteinerLehmus Theorem (Any triangle that has two equal angle bisectors each measured from a polygon vertex to the opposite sides is an isosceles triangle.)
And there you have it. Case solved.
Given:
AF = DB
AF bisects <CAB
DB bisects <CBA
Step 1:
AF = DB
Reason:
Given
Step 2:
Segment AF bisects angle CAB and extends to the side opposite it (Side CB)
Reason:
Given in diagram
Step 3:
Segment DB bisects angle CBA and extends to the side opposite it (Side CA)
Reason:
Given in diagram
Step 4:
Triangle ABC is isosceles.
Reason:
SteinerLehmus Theorem (Any triangle that has two equal angle bisectors each measured from a polygon vertex to the opposite sides is an isosceles triangle.)
And there you have it. Case solved.
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From reading the section about this problem in Geometry Revisited, I get the impression that there is no direct way of proving it. Coxeter uses the SteinerLehmus theorem and says that
Besides, reductio ad absurdum is a valid mathematical way of proving something.each [allegedly direct proof] is really an indirect proof in disguise.
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Hi, guys!
look at this:
http://www.mathematik.unibielefeld.de/ ... nerlehmus
especially this one:
An algebraic proof: [Baptist], [Specht, D.8, D.68, D.69, D.70]
Thm of Stewart: Length of a traversal CP
t^2 = m a^2 / c + n b^2 /c  m n = (m a^2 + n b^2)/(m + n)  n m
For the angle bisector at point C we know the ratio m : n = b : a.
With c=m+n we can deduce m = bc/(a+b) and n = ac/(a+b) and
w_c^2 = ab (1  c^2/(a+b)^2) = ab(a+b+c)(a+bc)/(a+b)^2
Now we have equal angle bisectors w_a = w_b that is
0 = w_a^2  w_b^2
= c(a+b+c)( b(b+ca)/(b+c)^2  a(a+cb)/(a+c)^2 )
= c(a+b+c)(ba)( (a+b+c)(ab+c^2) + 2abc )/( (b+c)^2 (a+c)^2 )
Only the third factor be zero for positive a, b, c.
Therefore we have a = b.
look at this:
http://www.mathematik.unibielefeld.de/ ... nerlehmus
especially this one:
An algebraic proof: [Baptist], [Specht, D.8, D.68, D.69, D.70]
Thm of Stewart: Length of a traversal CP
Code: Select all
C
/ / \
b / /t \ a
/ / \
APB
m n
For the angle bisector at point C we know the ratio m : n = b : a.
With c=m+n we can deduce m = bc/(a+b) and n = ac/(a+b) and
w_c^2 = ab (1  c^2/(a+b)^2) = ab(a+b+c)(a+bc)/(a+b)^2
Now we have equal angle bisectors w_a = w_b that is
0 = w_a^2  w_b^2
= c(a+b+c)( b(b+ca)/(b+c)^2  a(a+cb)/(a+c)^2 )
= c(a+b+c)(ba)( (a+b+c)(ab+c^2) + 2abc )/( (b+c)^2 (a+c)^2 )
Only the third factor be zero for positive a, b, c.
Therefore we have a = b.

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Any proof of the theorem would appear to be a challenge. And any legit proof is equally valid, no matter what it uses to get there. Why the fixation on the Law of Sines?threewood14 wrote:I think the challenge is using the Law of Sines...
I, Lex Llama, super genius, will one day rule this planet! And then you'll rue the day you messed with me, you damned dirty apes!

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