# [SOLVED] Geometry

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• Feb 8th 2009, 05:27 PM
kgk
[SOLVED] Geometry
In triangleABC
D is a point on BA such that BD:DA = 1:2
E is a point on CB such that CE:EB = 1:4
DC and AE intersect at F
I am suppose to find CF:FD(express the ratio in terms of two positive whole numbers) Help!
• Feb 9th 2009, 02:27 AM
Grandad
Geometry
Hello kgk
Quote:

Originally Posted by kgk
In triangleABC
D is a point on BA such that BD:DA = 1:2
E is a point on CB such that CE:EB = 1:4
DC and AE intersect at F
I am suppose to find CF:FD(express the ratio in terms of two positive whole numbers) Help!

You can solve this using ratios and position vectors. See my post at http://www.mathhelpforum.com/math-he...tml#post262231 for the basic formula.

In your problem, if we denote the position vectors of A, B, ... by $\vec{a}, \vec{b}, ...,$ then

$\vec{d} = \frac{1}{3}\vec{a} +\frac{2}{3}\vec{b}$

$\vec{e} = \frac{4}{5}\vec{c} + \frac{1}{5}\vec{b}$

If we suppose that F divides CD in the ratio 1:n, and EA in the ratio 1:m, then have two ways of expressing $\vec{f}$.

Using CF:CD = 1:n, $\vec{f} = \frac{n}{n+1}\vec{c} + \frac{1}{n+1}\vec{d}$

$=\frac{1}{3(n+1)}\vec{a}+ \frac{2}{3(n+1)}\vec{b} +\frac{n}{n+1}\vec{c}$

Using EF:FA = 1:m, $\vec{f} = \frac{m}{m+1}\vec{e}+\frac{1}{m+1}\vec{a}$

$= \frac{1}{m+1}\vec{a}+\frac{m}{5(m+1)}\vec{b}+\frac {4m}{5(m+1)}\vec{c}$

If we equate these two expressions for $\vec{f}$ and compare coefficients of $\vec{a}$, $\vec{b}$ and $\vec{c}$, we get:

$\frac{1}{3(n+1)}=\frac{1}{m+1}$ (1)

$\frac{2}{3(n+1)}=\frac{m}{5(m+1)}$ (2)

$\frac{n}{n+1} =\frac{4m}{5(m+1)}$ (3)

From (2) and (3): $\frac{n}{n+1} = \frac{8}{3(n+1)}$

$\Rightarrow n = \frac{8}{3}$

Check: $\Rightarrow m = 10$. Subst into (1): $\frac{1}{3\times \frac{11}{3}} = \frac{1}{11}$

So the ratio CF:FD = $1:\frac{8}{3}$ = 3:8.

Grandad