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. Given: 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 one. 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.  In the same way you can find BE*(a+c)*sin(0.5*B) = 2O.  From  =  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))  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  is positive. However cos(0.5*B)<cos(0.5*A) . 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)  Combining  and  results in: b*sin(0.5*A) < a*sin(0.5*B) So the lefthand side of  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 http://mathforum.org/dr.math/
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) = angle(B). 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) = angle(BFE). 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 way. 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 problem. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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