Find AX in the diagram if CX bisects angle ACB.

And then there's a diagram below with triangle ABC, where X is on line C, BC=45, AC=21, and BX=30.

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1. take a look at the Angle Bisector Theorem

AX/BX = CA/CB

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2. This is really just about the base toes of a triangle, namely triangle ABC. This is the complete solution for this particular problem:

The Angle Bisector Theorem tells us that $\frac{AC}{AX}=\frac{BC}{BX}$so $AX=\frac{AC\cdot BX}{BC}=\frac{21\cdot30}{45}=\boxed{14}.$

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