# Math Help - 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 $a.\sqrt[3]{2}$

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

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.