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Bisectors in Triangles!!! Pls Help!

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In triangle ABC, angle ABC = 90 degrees, and point D lies on segment BC such that AD is an angle bisector. If AB = 105 and BD = 42, then find AC.

MIRB15  Jul 8, 2017
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Because angle BAC is bisected  by AD, we have that

BD / DC  =  AB / AC       so

42 / DC = 105 /AC     which implies that

105 * DC  =  42AC

DC  =  (42/105)AC =  (2/5) AC  =  .4AC

So....using the Pythagorean Theorem we have that

AB^2  + BC^2  = AC^2

AB^2  + ( BD + DC)^2  = AC^2

105^2  + ( 42 + .4AC) = AC^2   simplify

105^2  + 42^2 + 33.6AC + .16AC^2  =  AC^2

12789 + 33.6AC + .16AC^2  =  AC^2    subtract the left side from both sides

.84AC^2   - 33.6AC - 12789 =  0

Using the quadratic formula will give us the positive solution for AC  =  145

CPhill  Jul 8, 2017

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