Find Angle ACBDate: 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 computation. 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) and 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) and 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 -a^2+b^2+c^2 cos(A) = ------------ 2bc 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): -2bac^2(b^2+ab+a^2-c^2) ----------------------- = 0 (a+b)^2(a+c)(b+c) 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 http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/