Without loss of generality, we assume $\\triangle ABC$ is a right triangle with legs $AB=20$ and $AC=11$. Constructing the rest of the diagram, we have this: <\/p>\n
![\"\"](\"http:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2021\/07\/diagram-4.png\")
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Using the angle bisector theorem, we have $\\dfrac{AB}{AC}=\\dfrac{CD}{DB}=\\dfrac{20}{11}.$ We now can set up mass points, with $\\text{mass}(B)=11$ and $\\text{mass}(C)=20.$ Then $\\text{mass}(D)=\\text{mass}(B)+\\text{mass}(C)=20+11=31.$ Because $M$ is the midpoint of $AD$, then $\\text{mass}(A)=\\text{mass}(D)=31.$ Therefore, the ratio $\\dfrac{CP}{PA}=\\dfrac{\\text{mass}(C)}{\\text{mass}(A)}=\\boxed{\\dfrac{20}{13}}.$<\/p>\n\n\n
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(AIME 2011) In triangle $ABC, AB=\\dfrac{20}{11}AC.$ The angle bisector of $\\angle A$ intersects $BC$ at point $D$, and point $M$ is the midpoint of $AD$. Let $P$ be the point of intersection of $AC$ and $BM$. Find $\\dfrac{CP}{PA}$. SolutionWithout loss … Continue reading