Isosceles Triangle

well, the bisectors are part of a family of isosceles triangles on that base, but with lowering height. As you squeeze the top down, keeping angles constant it will pass through the bisected position = simple prrof

Let the equal lengths be x

Drop a perpendicular (height h) from the intersection to the base.

Now sine of one half angle (h/x) equals the sine of the other half angle (h/x): so the angles themselves must be equal.

Make sense?

Since the bisectors of the base angles are equal in length and share the base between them, it follows that the sine of the two halved base angles must be equal. So the sine of the base angles and the base angles themselves must also be equal.

Proof Police Here! Put your hands up! Using trig functions is illegal in geometric proofs.

Even if trig functions were legal for the proof, you would first have to prove that the half angles are equal. The "sine of one half angle equals the sine of the other half angle" only if you have proved that the half angles are the same, which you have not.

Also, remember that the lines of length X extend beyond your intersection. Therefore, the right triangles to which you would apply trig (if you could) do not include a side of x length. The sines you mention are really h/a and h/b. You have not proven that a and b are equal.

Remember how obsessive-compulsive your geometry teacher was?

OK Ken,

I have to hold my hands up to several things. First, I misunderstood the question: in the original post Yesyen did not specify where the lines ended, and, I assumed they ended where they met each other. Second, I used trig. because I thought it would be more accessible to most people who visit the site; I could have as easily invoked congruent triangles. Third, my maths teachers were very well balanced chaps I'm afraid it's me who's "obsessive compulsive", and NOW we have a real problem.

Ken and Yesyen,

This has taken quite a long time (not just elapsed), so please have a look to see if you can spot any errors or omissions.

i.) First observe that circles on the same (or equal) chords which subtend different acute angles at the circumference obey the rule that the smaller circle subtends the bigger angle and vice versa.

Start by assuming that a triangle exists which is not isosceles and has equal bisectors. Start with the case where both "bottom" angles are acute and that without Loss Of Generality A is smaller than B.

Notice that D is lower than E (because AD = BE and the half angle at A is less than the half angle at B)

Let CD/DB = a, and, CE/EA = b

Draw the four circumcircles ADC, ADB, BEC and BEA with radii r1, r2, r3 and r4 (I have only shown the first two here)

Now r1/r2 = a (drawn from chords with this ratio and subtending equal angle A/2 at their circumference)

similarly r3/r4 =b

but: r1 = r3 (drawn on equal chords subtending the same angle C)

therefore r2*a =r4*b
therefore a/b = r4/r2
but a/b > 1 , and, because of i.) r4/r2 < 1

contradiction

For the case where B is obtuse notice that the angle subtended on the opposite side of chord AD equals A + C (opposite angles of a cyclic quad, and, angle of a triangle) which is also greater than A.

If you place the point of a pair of compasses where the bisectors meet and scribe a circle of radius equal to the bisector length, then the circle must pass through the opposite end of both bisectors and the base will form a chord. A segment will be formed by the chord (or triangle base) and the portion of the circle away from the bisectors.

I have included a piccy (if it comes out...I haven't tried this before) from wikipedia which is a great source of information (though strangely I didn't bother looking up your original question...I wonder what is says!). You can see the isosceles triangle formed of R,s,R

Yet another thought, to refresh memories: There is a big difference between proving a: that the bisectors of the base angles of an isosceles triangle are always equal (an easy proof), and

b: that if the bisectors are equal length, the triangle is isosceles.

If you prove that all tigers are animals, you have not necessarily proven that all animals are tigers.

It is obvious that only in an isosceles triangle the bisectors of base angles will be equal. But, how to prove it (geometrically)?

Let us take any triangle as shown and prove ABC is isosceles.

What are given in triangle ABC: AD and BE are bisectors of angle A and angle B respectively. AD = BE

In triangles: ABD and ABE, AB is common, AD = BE.

This is just not enough to say ABD and ABE are congruent.

Two triangles can be thought of having same base and sides equal, need not have the included angles equal, therefore need not have the trigonometrical relations also equal.

Here's what I think:

In the diagram above, Triangle ABC is an Isosceles Triangle. By definition that means:

AB=AC

ABC=ACB

When we draw bisectors through ABC and ACB we split AC and AB in half (by definition). The midpoint of AC being Q and the midpoint of AB being M.

We now have two new triangles, BQC and CMB. Both have a common side in BC. From above we know that QC = MB. We also know that MBC = QCB.

So the two triangles BQC and CMB are equal by Side Angle Side (SAS). If the two triangles are equal, then CM = BQ. So the bisectors of an Isoceles triangle are equal.

This proof above only proves that bisectors are equal in Isoceles Triangles. It doesn't show that bisectors are not equal in Irregular Triangles.

Here is a great page with a bunch of proofs.

Roger: A couple things to think about:

1. You have actually started with that which is to be proved, and ended where you should start. If you look at yesyen's second post, the challenge is clarified.

2. Also M and Q are not necessarily the midpoints. If you redrew your triangle so it was taller, you would see that M and Q would then be below the midpoint.

This would be easier if we were younger.

Ken,

1. Yes, you're right. I tried to indicate that in the bottom of my post.

2. Yes, I think you're right. Still, since the angles being bisected are the same, the line segments they produce opposing them will be equal, which still allows us solve for the bisectors by comparing two like triangles. The next step would be to show that if the base angles are different then the bisectors are not equal.

If we prove:

1. That equal base angles produce equal bisectors

2. Unequal base angles produce unequal bisectors

I think it proves that:

Equal bisectors indicate equal base angles (thus Isosceles Triangle)

I don't know how else to prove it. I hope someone finds the proof online and puts it up, I'd like to see it solved.

Roger:

Here's a simple way to prove it: ALL triangles are isosceles!

See:

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

I think, but am not sure, that the proof can be direct, without having to prove that triangles without equal length bisectors are not isoscles. If I have some time later, I'll play around, and maybe call my son in, who has taken geometry more recently.

Why not we post this puzzle to Jim Loy (www,jimloy.com)?

I spent some time drawing triangles, and couldn't come up with a proof that doesn't start with the isosceles triangle. So I threw up my hands and searched on line. I found these links which indicate that the proof is not easy. So I think we can all be excused for saying OK, enough of this silliness, we have things to do, places to be, lives to live.

Turns out Jim Loy has already given it some thought.

sir my teacher has given me the same question and I have worked on it for ages but I still havn't managed to find the answer would you please give it to me

I would be thakfull if you send it to this adress vahidowen2003@yahoo.co.uk

The solution:

http://forogeometras.com/index.php?topic=1272.msg3570;topicseen#new