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# Hard geometry! PLS Help ASAP!

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In triangle ABC, AB = 9, BC = 12, AC = 15, and CD is the angle bisector. Find the length of CD.

MIRB15  Jul 7, 2017
edited by MIRB15  Jul 7, 2017
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#1
+75016
+1

I'm assuming that CD  is drawn to side AB

We have a Pythagorean right triangle  9 -12 - 15

Angle  ACB can be expressed  using the inverse tangent

arctangent (9/12)  =  ACB

And DCB =  .5arctangent (9/12)

Likewise, angle BDC = 90 - DCB = 90 - .5*arctangent (9/12)

So....using the Law of  Sines, we have that

CD/ sin DBC  =  BC /  sin DCB       →  DBC  = 90  →  sin 90  =  1

CD  =  12 / sin  ( 90 - .5 *arctangent (9/12) )   ≈  12.65

Here's a pic

CPhill  Jul 7, 2017
#2
+75016
+1

Here's another way using just Geometry

Since   ACB is bisected......then  DC divides  AB   in a ratio of  12 /15  = BD/ DA

So.....BD  =  (12/27) * 9  =   (4/9)*9  = 4

Using the Pythagorean Theorem

CD  = √[ BD^2  + BC^2] =  √ [ 4^2  + 12^2 ]  =  √ [ 160 ]  ≈  12.65

CPhill  Jul 7, 2017
edited by CPhill  Jul 7, 2017

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