Question

in any triangle ABC ,if a/cosA =b/cosB,then prove that is an isosceles

Answer 1

Using the Law of cosines for any triangle (here is very important, we are not assuming the type of that triangle, Law of cosines applies to any triangle), so in the triangle ABC we have: cosA = (b2 + c2 - a2)/2bc So the left side of that equation will be:   cosA/b  = (b2+ c2 - a2)/ (2*b2*c)      (one more b in the denominator ) As the same reason, the right side of that equation will be:  cosB/a =  (a2 +c2 - b2)/ (2*a2*c) So if  cosA/b = cosB/a  We will have that  (b2+ c2 - a2)/ (2*b2*c)  =  (a2 +c2 - b2)/ (2*a2*c)  Then we simplify that equation as following steps: (b2+ c2 - a2)/ b2 = (a2 +c2 - b2)/ a2         (both sides have the 2c in the denominator, so 2c eliminated) then in either of the two sides, we divide the denominator into the bracket, then we get: 1+ c2/b2 - a2/b2  =  1 + c2/a2 - b2/a2 Then we eliminate the same part of both sides, which is   1  ,we will get an equation as: c2/b2 - a2/b2= c2/a2 - b2/a2 We multiply a2b2 at each side (because neither a nor b is zero for they are sides of triangle), then we get c2a2 - a4 = c2b2 - b4
 move the c2 to same sides, we will get: c2(a2-b2) = a4 - b4 = (a2 + b2) (a2 - b2) Ok, till now we get an equation as : c2 (a2-b2) = (a2+b2) (a2-b2)      -------------------------- EQUATION_ONE Now we have to discuss the different situation: 1. If a2-b2 = 0, which means a = b (because a and b are sides of triangle, so they are positive), then that EQUATION_ONE will be satisfied no matter c is what.  So that means the triangle will be isosceles triangle, as a=b. 2. If (a2-b2) is not 0, then we can eliminate the (a2-b2) in both sides of the EQUATION_ONE . Then we will have: c2=a2+b2 which means the triangle is a right angle triangle    In conclusion, if cosA/b = cosB/a, we can get c2 (a2-b2) = (a2+b2) (a2-b2) by using Law of cosines, and then we can get that either  c2=a2+b2  or a2-b2=0   can satisfy that equation above, which means the triangle is either right angle triangle or an isosceles triangle. 

Answer 2

the answer is given in the above image.