In a triangle ABC, AB = 2.3; BC = 3.6; AC = 5. The bisector of the VM was drawn; the segment of the MK is plelen AB

In a triangle ABC, AB = 2.3; BC = 3.6; AC = 5. The bisector of the VM was drawn; the segment of the MK is plelen AB as it is located on the BC. Find the lengths of the segments CM, MA, MK.

Given:
ABC – triangle
AB = 2.3
BC = 3.6
AC = 5.
BM – bisector
MK || AB.
To find:
CM – ?
MA -?
MK -?
Decision:
Consider ΔАВС:
MS = x
BM – bisector
AM / MS = AB / BC
5-x / x = 2.3 / 3.6
5-x / x = 23/36
180-36x = 23x
59x = 180
x = 180/59 = 3 3/59 = MC
MA = 5-3 3/59 = 156/59
ΔМКС is similar to ΔАВС, since MK || AB:
MK / AB = MS / AS
MK = AB * MS / AC = 2.3 * 180: 59/5 = 23 * 18/59 * 5 = 414/295 = 1 119/295