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 $CE?$
\(BD = k \sqrt{2}\)
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)
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