In $\triangle ABC$, point $E$ bisects $BC$ and point $D$ trisects $AB$. If $CD$ and $AE$ meet at $P$, determine $CP:PD$.

Question

In a triangle $ABC$, $E$ is the midpoint of $BC$ and $D$ is a point on $AB$ such that $AD : DB = 2 : 1$. If $CD$ and $AE$ intersect at $P$, determine the ratio $CP : PD$.

How do I solve it using only section formula and basic coordinate geometry. Without using straight lines or vectors to simplify.

Edit: I have taken the co-ordinates of vertices as $( x_a , y_a )$ so on.

I have assumed that $\overleftrightarrow{AE}$ is divided in the ratio $\sigma$:1 by P and $\overleftrightarrow{CD}$ is divided in the ratio $\lambda$:1 by P.

and I have got an equation for $x$ co-ordinate as $$\frac{\lambda(x_a + 2x_b) +3x_c}{3(\lambda + 1)} = \frac{\sigma(x_b +x_c) + 2x_a}{2(\sigma+1)}$$

I dont know what has to be done next. The answer is $3:2$

My attempt

Answer 1

Lets consider the $\triangle CDB$ which is cut by the line $AE$. According to Menelaus theorem: $$\begin{align}\frac{BE}{EC}\cdot \frac{CP}{PD}\cdot \frac{DA}{AB}&=1\\ \frac{CP}{PD}&=\frac{EC}{BE}\cdot \frac{AB}{AD}\\ &=1\cdot\left(\frac{2x+x}{2x}\right)\\ &=\frac{3}{2}\end{align}$$

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Alternate Method:

Draw a $DF$ parallel to $AE$. We have, by basic proportionality theorem in $\triangle BAE$.

$$\frac{AD}{BD}=\frac{FE}{BF}\implies \frac{2}{1}=\frac{FE}{BF}$$

$E$ is the midpoint of $BC$, so $EC=x+2x=3x$.

Similarly for $\triangle CDF$, $$\frac{CP}{PD}=\frac{EC}{FE}\implies\frac{CP}{PD}=\frac{3x}{2x}=\frac{3}{2}$$

Answer 2

If a triangle satisfies these conditions, the ratio $\frac{\overline{CP}}{\overline{PD}}$ will always be the same no matter what the triangle looks like.

So, to make the computing easier, you can just place $B$ at $(0,0)$, $A$ at $(0,3)$, $C$ at $(2,0)$, $D$ at $(0,1)$, and $E$ at $(1,0)$ for instance.

Now find the linear equations of $\overleftrightarrow{CD}$ and $\overleftrightarrow{AE}$, and then figure out the coordinates of $P$. At this point, you can solve this question with section formula.

Answer 3

In the diagram, mark the centroid "$G$" (say) on $AE$.
Now join $D$ and $G$.
you will notice that $\triangle{PEC} \sim \triangle{PGD}.$

now, $\frac{BE}{GD} = \frac{3}{2}$ ( using Thales Theorem in $\triangle{ADG}$ & $\triangle{ABE}$ ).

Thus in $\triangle{PEC}$ and $\triangle{PGD}$,

we get that $\frac{CP}{PD} = \frac{CE}{BD} = \frac{3}{2}$ ($\because$ CE=BE)