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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
+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

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 =  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)

CPhill  Nov 19, 2018
#2
+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

CPhill  Nov 19, 2018