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Find Angle ACB

Date: 04/20/2002 at 01:33:40
From: Bennette
Subject: Triangles 

Let A', B' and C' be points on triangle ABC such that AA' BB' CC' are 
angle bisectors. Suppose angle A'C'B' = 90 degrees. Find angle ACB.

No idea at all. please help

Date: 04/21/2002 at 16:33:42
From: Doctor Floor
Subject: Re: Triangles 

Hi, Bennette,

Thanks for your question.

Angle ACB must be equal to 120 degrees. I only found that by ugly 

The angle bisector AA' divides BC(=a) in the ratio BA':A'C = AB:AC = 
c:b. This can be seen from the fact that the distances from a point 
on AA' to AB and AC are equal. From that the areas of triangles AA'B 
and AA'C are in the ratio of AB and AC, from which we see that BA' 
and A'C must be in the same ratio, since on these bases the altitude 
from A is equal.

Similarly AB':B'C = c:a and BC':C'A = a:b.

This shows us that 

  BC' = c * a/(a+b) = ac/(a+b) 


  BA' = a* c/(b+c) = ac/(b+c).

By the Law of Cosines this gives

                              1           1       2cos(B)
  b'^2 = (A'C')^2 = a^2c^2*(-------  + ------- - --------- ).
                            (a+b)^2    (b+c)^2   (a+b)(b+c)

In a similar fashion:

                              1           1       2cos(A)
  a'^2 = (B'C')^2 = b^2c^2*(-------  + ------- - --------- ),
                            (a+b)^2    (a+c)^2   (a+b)(a+c)


                              1           1       2cos(C)
  c'^2 = (A'B')^2 = a^2b^2*(-------  + ------- - --------- ).
                            (a+c)^2    (b+c)^2   (a+c)(b+c)

Now we can substitute the expressions like

 cos(A) = ------------

which we acquire from the Law of Cosines. Then the fact that angles 
A'C'B' is equal to 90 degrees, gives us c'^2 - a'^2 - b'^2 = 0. This 
gives us after the three substitutions and simplifications (this is 
quite a job):

  ----------------------- = 0

Since the other factors cannot yield 0, this leaves that

  b^2+ab+a^2-c^2 = 0

and thus

  c^2 = a^2+b^2+ab

from which we see, again using the Law of Cosines, that cos(C) must be 
equal to -1/2, and thus C must be equal to 120 degrees.

If you have more questions, just write back.

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

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