Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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

share|cite|improve this answer

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.

share|cite|improve this answer

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

Referring to the figure,here

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.