# Exam Geometry Question

• Apr 22nd 2009, 05:38 PM
xwrathbringerx
Exam Geometry Question
In the diagram, triangle ABC is equilateral, the length of the sides being a. Base line AB is produced in both directions to D and E respectively such that AD = AB = a, while E is the intersection of line AB produced and the line CFE. Line BF is parallel to line DC and the length of line EF is again a.

Show that the length x of line CD is $\displaystyle a.\sqrt[3]{2}$

• Apr 25th 2009, 03:23 AM
Hello xwrathbringerx
Quote:

Originally Posted by xwrathbringerx
In the diagram, triangle ABC is equilateral, the length of the sides being a. Base line AB is produced in both directions to D and E respectively such that AD = AB = a, while E is the intersection of line AB produced and the line CFE. Line BF is parallel to line DC and the length of line EF is again a.

Show that the length x of line CD is $\displaystyle a.\sqrt[3]{2}$

I take it you mean the line CF.

Draw P as the foot of the perpendicular from C to AB. Then in triangle CPE

$\displaystyle PE^2 = (x+a)^2-CP^2$

$\displaystyle = (x+a)^2-\tfrac34a^2$, using triangle CBP

$\displaystyle = x^2+2ax+\tfrac14a^2$

$\displaystyle \Rightarrow BE = \sqrt{x^2+2ax+\tfrac14a^2} - \tfrac12a$

Now triangles CED and FEB are similar because CD is parallel to FB

So $\displaystyle \frac{CF}{FE} = \frac{DB }{BE}$

$\displaystyle \Rightarrow \frac{x}{a}= \frac{2a}{BE}$

$\displaystyle \Rightarrow \sqrt{x^2+2ax+\tfrac14a^2} - \tfrac12a = \frac{2a^2}{x}$

$\displaystyle \Rightarrow 4(x^2+2ax+\tfrac14a^2) = \left(a + \frac{4a^2}{x}\right)^2$

$\displaystyle = a^2 + \frac{8a^3}{x}+\frac{16a^4}{x^2}$

$\displaystyle \Rightarrow 4x^4+8ax^3+a^2x^2= a^2x^2+8a^3x + 16a^4$

$\displaystyle \Rightarrow 4x^4 + 8ax^3 - 8a^3x -16a^4=0$

$\displaystyle \Rightarrow x^4 +2ax^3 -2a^3x-4a^4=0$

$\displaystyle \Rightarrow (x^3-2a^3)(x+2a)=0$

$\displaystyle \Rightarrow x = \sqrt[3]{2}a$, taking the positive root.