# Ratio of two unknown line segments using known side

Point P lies on the side AB of triangle ABC, and the ratio between AP and PB is 1 : 3

The median line from vertex A intersects line segment CP at point Q and side BC at point D

What is the ratio between AQ and QD

The answer should be 2 : 3 but I can't reach it! A vector approach would be optimal and greatly appreciated. Thanks in advance!

• Hint: use Menelaus' theorem for $\,\triangle ABD\,$ and transversal $\,CP\,$ – dxiv Apr 5 '18 at 23:15
• Lovely hint @dxiv! I just learned something new. Thanks! – larryK Apr 5 '18 at 23:48

(The most direct solution would use Menelaus' theorem, as pointed out in a comment already. The following is an alternative way to solve it, per the "vector approach" part of OP's question.)

By construction:

\begin{cases} \begin{align} \overrightarrow{AP} &= \frac{\overrightarrow{AB}}{4} \\[5px] \overrightarrow{AD} &= \frac{\overrightarrow{AB}+\overrightarrow{AC}}{2} \end{align} \end{cases} \tag{1}

Let $\,\color{blue}{\dfrac{AQ}{QD} = \lambda} \iff \dfrac{AQ}{AD} = \dfrac{\lambda}{\lambda+1} \,$, then:

$$\overrightarrow{AQ} = \dfrac{\lambda \, \overrightarrow{AD}}{\lambda+1} \tag{2}$$

Let $\,\dfrac{CQ}{QP} = \mu\,$, then:

$$\overrightarrow{AQ} = \overrightarrow{AC}+\overrightarrow{CQ}=\overrightarrow{AC}+\dfrac{\mu\,\overrightarrow{CP}}{\mu+1}=\overrightarrow{AC}+\dfrac{\mu\,\big(\overrightarrow{AP}-\overrightarrow{AC}\big)}{\mu + 1}=\dfrac{\overrightarrow{AC}+\mu\,\overrightarrow{AP}}{\mu+1} \tag{3}$$

Equating $\,(2)\,$ and $\,(3)\,$, then using $\,(1)\,$:

$$\dfrac{\lambda \, \overrightarrow{AD}}{\lambda+1} = \dfrac{\overrightarrow{AC}+\mu\,\overrightarrow{AP}}{\mu+1} \;\;\iff\;\; \frac{\lambda}{\lambda+1} \cdot \frac{\overrightarrow{AB}+\overrightarrow{AC}}{2} = \frac{1}{\mu+1}\cdot \overrightarrow{AC}+\frac{\mu}{\mu+1}\cdot\frac{\overrightarrow{AB}}{4} \tag{4}$$

Since $\,\overrightarrow{AB}, \overrightarrow{AC}\,$ are linearly independent, the component-wise coefficients must match in $\,(4)\,$:

\begin{cases} \begin{align} \frac{\lambda}{2(\lambda+1)} &= \frac{\mu}{4(\mu+1)} \\[5px] \frac{\lambda}{2(\lambda+1)} &= \frac{1}{\mu+1} \\[5px] \end{align} \end{cases}

It then easily follows that $\,\mu=4\,$ and $\,\color{blue}{\lambda=\dfrac{2}{3}}\,$.

• Is this derivation equivalent to proving Menelaus' theorem? – marty cohen Apr 6 '18 at 3:58
• @martycohen Not as written, but it could easily be made into a proof of Menelaus' or Ceva's theorems. – dxiv Apr 6 '18 at 4:37