In triangle ABC: if cosA/b = cosB/a proof that the triangle either right angle triangle or isosceles triangle

Question

Can you show me how to proof that this triangle in both situations as fast as you can

thanx in advance

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

if isosceles with angle A = angle B, then a = b because sides opposite equal angles are equal

 

Therefore cosA/b = cosB/b = cosB /a

 

if right triangle, C = 90 degrees

 

cos A = b/c or cos A/b = 1/c

 

cos B = a/c or cos B /a = 1/c

 

Both equal 1/c, therefore cosA/b = cosB/a