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Find the ratio of the area of \(\triangle BCX\) to the area of \(\triangle ACX\) in the diagram if CX bisects \(\angle ACB\). Express your answer as a common fraction.

Guest Dec 14, 2018

#1**+1 **

Let AX = M

Let BX = 24 - M

Since ACB is bisected, we have this relationship

M / 30 = (24 - M) / 27 cross-multiply

27M = 30( 24 - M) simplify

27M = 720 - 30M add 30M to both side

57M = 720

M = 720 / 57 = 240 / 19 = AX

Then 24 - M = 24 - 240/19 = 216/19 = BX

Since each triangle is under the same height, the ratio of their areas is just the ratio of their bases

So

area BCX / Area ACX = ( 216 /19) / (240 / 19) = 216 /240 = 9 / 10

CPhill Dec 14, 2018