Ratio of side length of triangle?

Question

In triangle ABC, we choose a point D at AB such that the length of AD=1/2 AB, and point E at AC such that AE=3EC. F is intersection point of CD and BE. What is the ratio of CF/FD and BF/FE?

Answer 1

Let $CF:FD = (1-p):p$, $BF:FE = (1-q):q$.

Let $\vec{AB}$ be $2\mathbf{b}$, $\vec{AC}$ be $4\mathbf{c}$. Then $\vec{AD} = \mathbf{b}$ and $\vec{AE} = 3\mathbf{c}$.

By ratio, $$\begin{align} \vec{AF} =& p\vec{AC} + (1-p)\vec{AD} = 4p\mathbf{c}+(1-p)\mathbf{b}\\ \vec{AF} =& q\vec{AB} + (1-q)\vec{AE} = 2q\mathbf{b}+3(1-q)\mathbf{c}\\ \end{align}$$

Matching the coefficients of $\mathbf{b}$ and $\mathbf{c}$, $$\begin{align} &\begin{cases} 1-p = 2q\\ 4p = 3(1-q) \end{cases}\\ &\begin{cases} p + 2q = 1\\ 4p + 3q = 3 \end{cases}\\ &\begin{cases} p = \frac{3}{5}\\ q = \frac{1}{5} \end{cases}\\ \end{align}$$

Therefore, $CF:FD = 2:3$, $BF:FE = 4:1$.

Answer 2

Here is a way I was taught at school. It helps to draw a diagram

Let the vector $AB=\vec b$ and $AC= \vec c$ and choose $A$ as the origin.

[We use that if $OP=\vec p, OQ=\vec q$ then $PQ=\vec q - \vec p$ and a point on the line $PQ$ is at $\vec p +\alpha (\vec q -\vec p)$ for some value of $\alpha$]

$D$ is then at position $\frac 12 \vec b$ and $E$ at $\frac 14 \vec c$

The line $CD$ is then $\vec c+\lambda (\frac 12 \vec b-\vec c)=(1-\lambda)\vec c+\frac{\lambda}2\vec b$

The line $BE$ is $\vec b+\mu (\frac 14 \vec c-\vec b)=(1-\mu)\vec b+\frac {\mu}4\vec c$

F is the point where these two lines meet, which involves equating coefficients of $\vec b, \vec c$ - two equations in two unknowns. Then check how $\lambda, \mu$ are related to the ratios you need.

Answer 3

This can be solved by geometric method – [areas are prop. to bases if altitudes are the same.]

Referring to the figure,

Let the areas of triangle $\triangle DEF$ and $\triangle BCF$ be p and q respectively. Then,

{1} $= 3p + 3q$

{2} $= 2p + 3q$

{3} $= \frac {2(3p + 3q)} {3} – q = … = 2p + q$

Therefore, $\frac {2p + 3q} {2p + q} = \frac {p}{q}$

Let $\frac {p}{q}= $. After simplification, we have $2r^2 – r – 3 = 0$

From which, we get $CF : FD = q : p = 2 : 3$

The other ratio can be found similarly.