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Find Lengths of Sides of Triangle

Date: 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/ 
Associated Topics:
High School Triangles and Other Polygons

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