Find Lengths of Sides of TriangleDate: 04/20/2002 at 01:32:19 From: Bennette Subject: Triangle geometry Let ABC be a right-angled triangle with angle C = 90 degrees. Let the bisectors of angle A and angle B intersect BC and CA at D and E respectively. Given that CD=9 and CE=8, find the lengths of the sides of ABC. I thought of using similar triangles, but can't prove that triangle BCE and ACD and ABC are similar. Date: 04/22/2002 at 09:54:15 From: Doctor Floor Subject: Re: Triangle geometry Hi, Bennette, Thanks for your question. We use notation a = BC, b = AC and c = AB. Note that CD:DB = b:c. This can be seen from the fact that the distances from any point on AD to AB and AC are equal, as they are from the center of the inscribed circle, which lies on AD. From that the areas of triangles ADC and ADB are in the ratio of b and c (having equal heights from b and c respectively), from which we see that BD and DC must be in the same ratio, since on these bases the altitude from A is equal. So we have CD 9 b -- = --- = - ....[1], DB a-9 c and thus (a-9)b = 9c ab = 9b+9c........[2]. In a similar fashion we find CE 8 a -- = --- = -.....[3], EA b-8 c and thus a(b-8)=8c ab = 8a+8c.......[4]. Combining [2] and [4] gives 8a+8c = 9b+9c c+9b a = ---- . 8 We can substitute this into a^2 + b^2 = c^2, giving c^2 + 18bc + 81b^2 ------------------ + b^2 = c^2 64 c^2 + 18bc + 81b^2 + 64b^2 = 64c^2 145b^2 + 18bc - 63 c^2 = 0 145(b/c)^2 + 18(b/c) - 63 = 0. From this quadratic equation we find that, since b/c must be positive, b/c = 0.6. Substitution into [1] yields that a = 24. What's more, we see that sin(<A) = b/c = 0.6, and this gives us without much trouble that cos(<A) = a/c = sqrt(1 - sin^2(<A)) = sqrt(1 - (0.6)^2) = 0.8. And thus we find, knowing a, that c = 30 and consequently b = 18. 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- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/