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Steiner-Lehmus Theorem

Date: 01/28/99 at 17:36:05
From: Caitlin McIntosh
Subject: Geometry proof

I have a proof about a triangle that I just can't figure out.

The triangle is ABC.  there are two angle bisectors from A and B.
The four angles formed are 1, 2, 3, and 4.

     angle 1 is congruent to angle 2
     angle 3 is congruent to angle 4
     the two bisectors are congruent

Prove: the triangle is isosceles

Date: 01/29/99 at 02:48:32
From: Doctor Floor
Subject: Re: Geometry proof

Hi Caitlin,

Thanks for your question.

The theorem you ask for is known as the Steiner-Lehmus theorem, and it 
is not an easy one. However, several proofs are known. I'll give you 

Let ABC be a triangle and let AD = BE be the (internal) angle 
bisectors of A and B. For the sides I use the names a:BC, b:AC and 
c:AB. For the area of  ABC I use O.

The area of triangle ADB is 0.5*c*AD*sin(0.5*A).
The area of triangle ADC is 0.5*b*AD*sin(0.5*A).

Triangles ADB and ADC add up to ABC, so AD*(b+c)*sin(0.5*A) = 2O. [1]

In the same way you can find BE*(a+c)*sin(0.5*B) = 2O. [2]

From [1] = [2] and AD = BE we derive:

             (b+c)*sin(0.5*A) = (a+c)*sin(0.5*B)
  b*sin(0.5*A) - a*sin(0.5*B) = c(sin(0.5*B)-sin(0.5*A)) [3]

Now suppose that angle B > angle A.

Then sin(0.5*B)>sin(0.5*A) (both half angles are < 90 deg), so the 
righthand side of [3] is positive.

However cos(0.5*B)<cos(0.5*A) [4].
From the law of sines we find b*sin(A) = a*sin(B). Using that 
sin(A) = 2*sin(0.5*A)*cos(0.5*A) and the same for sin(B) this gives:

  2*b*sin(0.5*A)*cos(0.5*A) = 2*a*sin(0.5*B)*cos(0.5*B)
    b*sin(0.5*A)*cos(0.5*A) = a*sin(0.5*B)*cos(0.5*B)  [5]

Combining [4] and [5] results in:

  b*sin(0.5*A) < a*sin(0.5*B)

So the lefthand side of [3] is negative and we have a contradiction. 
We can't have angle B > angle A. Likewise, we can't have angle 
A > angle B.

So angle A = angle B and triangle ABC is isosceles, so the 
Steiner-Lehmus theorem is proven.

I hope this helps. If you have a math question again, please send it 
to Dr. Math.

Best regards,  

- Doctor Floor, The Math Forum   

Date: 01/29/99 at 12:12:22
From: McIntosh, Caitlin
Subject: Re: Geometry proof

I understand... sort of. Could you explain it without trig?

Date: 02/01/99 at 07:56:07
From: Doctor Floor
Subject: Re: Geometry proof

Hi again Caitlin,

There are proofs of the Steiner-Lehmus theorem without trig. One of 
them is as follows.

Let ABC be the triangle, and let D be on BC and E on AC in such a way 
that AD and BE are the (internal) angle bisectors of A and B. The 
assumption is that AD = BE, and we want to prove that angle(A) = 

To prove this, construct a point F such that ADFE is a parallelogram: 
so AD = EF and AE = DF, and draw segment BF. 

We see that EF = AD, and since we assumed that AD = BE this means 
that EF = EB. So triangle BEF is isosceles, and thus angle(FBE) = 

Now suppose that angle(A) > angle(B). Then it is also true that 
angle(BAD) > angle(ABE), but since triangles BAD and ABE have two 
equal sides (AB = AB and BE = AD), we can conclude that BD > AE. On 
the other hand, angle(A) > angle(B) gives that angle(DAC) > 
angle(EBC), and since angle(DAC) = angle(EFD), it is also true that 
angle(EFD) > angle(EBC). We have already derived that angle(FBE) = 
angle(BFE), and this gives us that angle(DFB) < angle(DBF). In a 
triangle we know that a larger angle has a longer opposite side. 
Applying this to triangle DBF, we see that BD < DF, and since DF = AE 
we find that BD < AE. Thus assuming that angle(A) > angle(B) gives us 
that BD > AE as well as that BD < AE, and we have a contradiction, 
and our assumption cannot be true.

Assuming that angle(A) < angle(B) gives a contradiction in the same 

So angle(A) = angle(B), and we have proven the Steiner-Lehmus theorem!

I hope this explanation has made it easier for you to understand this 

Best regards,
- Doctor Floor, The Math Forum   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

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