Let $\triangle{ABC}$ be a triangle with area $120$ such that $AB=4AC$. Let $P$ be the intesection between median $CM$ and angle bisector $AD$.

Question

What is the area of $\triangle{PCD}$?

I have tried using mass points, and this is where it has gotten me:

$x=AC, 4x=AB.$

$\frac{x}{DC}=\frac{4x}{DB} \rightarrow DB(x)=4x(DC)$

Dividing both sides by $x$ yields:

$DB=4DC$.

From here on out, I figured that the mass of $A$ is 1, $B$ is 1, $M$ is 2, $C$ is 4, $D$ is 5, and $P$ is 6. However, I am still having trouble solving it.

May someone help me finish it off?

Answer 1

By the Angle bisector theorem, you have that $$\frac{MP}{PC} = \frac{AM}{AC} = 2.$$ Analogous $$\frac{BD}{DC} = \frac{AB}{AC} = 4.$$

By the first equality, we have $(AMP) = 40$ and $(APC)=20$. By the second equality, $(ABD)=96$, so $(PDC) = (ADC)-(APC) = 24-20=4. $