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1. In triangle $ABC$, $AB = 3$, $BC = 4$, $AC = 5$, and $BD$ is the angle bisector. If $BD = k \sqrt{2}$, then find $k$.

 

2.Let $C$ be a point not on line $AE$ and $D$ a point on line $AE$ such that $CD \perp AE.$ Meanwhile, $B$ is a point on line $CE$ such that $AB \perp CE.$ If $AB = 4,$ $CD = 8,$ and $AE = 5,$ then what is the length of $CE?$

Guest Nov 19, 2018
 #1
avatar+92785 
+1

\(BD = k \sqrt{2}\)

 

 

1. In triangle $ABC$, $AB = 3$, $BC = 4$, $AC = 5$, and $BD$ is the angle bisector. If $BD = k \sqrt{2}$, then find $k$.

 

A

 

3           5

 

B       4           C

 

 

The bisector will divide AC  into  a ratio  of 3 : 4   =  AD : CD

Therefore.....AD  =  (3/7)*5  =    15/7

 

The sine of BAC = 4/5

So  the cosine of BAC  =  √ [ 5^2 - 4^2 ] / 5  =   3/5

 

 

Using the Law of Cosines

 

BD^2   = AB^2 + AD^2  - 2 (AB * AD)cos BAC

 

BD^2 =  3^2  + ( 15/7)^2 - 2(3 * 15/7) (3/5)

 

BD^2  = 9 + 225/49 - 2(3 * 3 * 3) / 7

 

BD^2 =  9 +  225/49 - 54/ 7

 

BD^2  =   [ 441 + 225 - 378] / 49

 

BD^2  = 288 /49

 

BD = √288/ 7  =   (12/7) √2

 

So....   k  = (12/7)

 

 

 

cool cool cool

CPhill  Nov 19, 2018
 #2
avatar+92785 
+1

2.Let $C$ be a point not on line $AE$ and $D$ a point on line $AE$ such that $CD \perp AE.$ Meanwhile, $B$ is a point on line $CE$ such that $AB \perp CE.$ If $AB = 4,$ $CD = 8,$ and $AE = 5,$ then what is the length of $CE?$

 

 

                    C

              

                           B

 

              4

                     8

 

  A                D                                   E

 

                       5

 

We have two triangles ...  ABE  and CED

Angle ABE = Angle CDE

Angle BEA = Angle CED

 

Therefore.....by AA congruency....

 

Triangle ABE  is similar to Triangle  CDE

 

So

 

AE / AB   =  CE / CD

 

5 / 4   =  CE / 8        cross-multiply

 

8 * 5 /  4   = CE

 

40 / 4   =  CE

 

10  = CE

 

 

cool cool cool

CPhill  Nov 19, 2018

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