# Thread: Exam Geometry Question

1. ## 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 $a.\sqrt[3]{2}$

Please, any help?

2. Hello xwrathbringerx
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 $a.\sqrt[3]{2}$

Please, any help?
I take it you mean the line CF.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Grandad