Answer:
Certainly! Let's use the Side Splitter Theorem to find the ratio CP/CY.
Given that BX/XC = 3/5, let's extend BX to meet side AC at a point Z. Now, BX/XC = BZ/ZC = 3/5.
By the Side Splitter Theorem, we can write:
\[ \frac{AP}{PC} = \frac{AY}{YB} \]
Similarly, considering triangle ABC and the extension of BX:
\[ \frac{AZ}{ZC} = \frac{AY}{YB} \]
Now, we can set these two expressions equal to each other:
\[ \frac{AP}{PC} = \frac{AZ}{ZC} \]
Since \(BZ/ZC = 3/5\), we can substitute AZ/ZC with 3/5:
\[ \frac{AP}{PC} = \frac{3}{5} \]
Now, we know that \(AP/PC = 3/5\), and we want to find \(CP/CY\).
\[ \frac{AP}{PC} = \frac{CP + PA}{PC} = 1 + \frac{PA}{PC} = \frac{3}{5} \]
Now, solving for \(CP/PC\):
\[ \frac{CP}{CY + YB} = \frac{3}{5} \]
Since \(YB/YA = 1/2\), we can write \(YB = \frac{1}{2}YA\):
\[ \frac{CP}{CY + \frac{1}{2}YA} = \frac{3}{5} \]
Now, substituting \(YA = YC + CA\):
\[ \frac{CP}{CY + \frac{1}{2}(YC + CA)} = \frac{3}{5} \]
Simplifying:
\[ \frac{CP}{CY + \frac{1}{2}YC + \frac{1}{2}CA} = \frac{3}{5} \]
\[ \frac{CP}{\frac{3}{2}CY + \frac{1}{2}CA} = \frac{3}{5} \]
\[ \frac{CP}{3CY + CA} = \frac{3}{5} \]
Cross-multiplying:
\[ 5CP = 3(3CY + CA) \]
\[ 5CP = 9CY + 3CA \]
\[ 5CP = 3(3CY + CA) \]
\[ CP = \frac{3}{5}(3CY + CA) \]
Now, \(CP/CY\) can be expressed as:
\[ \frac{CP}{CY} = \frac{\frac{3}{5}(3CY + CA)}{CY} \]
\[ \frac{CP}{CY} = \frac{3}{5}\left(3 + \frac{CA}{CY}\right) \]
The value of \(CA/CY\) can be found using the fact that \(BX/XC = 3/5\):
\[ \frac{CA}{CY} = \frac{BA + AC}{CY} = \frac{BY + YC + AC}{CY} \]
\[ \frac{CA}{CY} = \frac{\frac{1}{2} + 1 + 3}{1} = \frac{9}{2} \]
Now, substitute this into the expression for \(CP/CY\):
\[ \frac{CP}{CY} = \frac{3}{5}\left(3 + \frac{CA}{CY}\right) \]
\[ \frac{CP}{CY} = \frac{3}{5}\left(3 + \frac{9}{2}\right) \]
\[ \frac{CP}{CY} = \frac{3}{5}\left(\frac{15}{2}\right) \]
\[ \frac{CP}{CY} = \frac{9}{2} \]
So, the ratio \(CP/CY\) is \(\frac{9}{2}\).
