# Geometrical problem

I have a problem to calculate At the given picture the line segments AB , AC , FE ,GH , GI are known I want to calculate the line segment ED Is this posible ? Thanks in advance Nikos Yes, it is possible for many cases:

This gives the relative position $\lambda \in [0,1]$: $$\lvert\overline{FE}\rvert = (1-\lambda) \lvert\overline{AB}\rvert + \lambda \lvert\overline{GH}\rvert = \lvert\overline{AB}\rvert + \lambda (\lvert\overline{GH}\rvert - \lvert\overline{AB}\rvert)$$

If we have a non-zero slope, $\lvert\overline{GH}\rvert \ne \lvert\overline{AB}\rvert$ we can solve for $\lambda$: $$\lambda = \frac{\lvert\overline{FE}\rvert - \lvert\overline{AB}\rvert}{\lvert\overline{GH}\rvert - \lvert\overline{AB}\rvert}$$

Then we have $$\lvert\overline{FD}\rvert = (1-\lambda) \lvert\overline{AC}\rvert + \lambda \lvert\overline{GI}\rvert$$

insert $\lambda$ and apply $$\lvert\overline{ED}\rvert = \lvert\overline{FD}\rvert - \lvert\overline{FE}\rvert$$

to get $$\lvert\overline{ED}\rvert = \lvert\overline{AC}\rvert + \frac{\lvert\overline{FE}\rvert - \lvert\overline{AB}\rvert}{\lvert\overline{GH}\rvert - \lvert\overline{AB}\rvert} (\lvert\overline{GI}\rvert- \lvert\overline{AC}\rvert) - \lvert\overline{FE}\rvert$$

There is also the boring case: $$\lvert\overline{GH}\rvert = \lvert\overline{AB}\rvert = \lvert\overline{FE}\rvert \wedge \lvert\overline{GI}\rvert = \lvert\overline{AC}\rvert$$ with solution $$\lvert\overline{ED}\rvert = \lvert\overline{AC}\rvert - \lvert\overline{AB}\rvert$$

I take it you mean that the lengths of the line segments and not their positions are given.

From the intercept theorem you get

$$\frac{\text{ED} - \text{BC}}{\text{AF}} = \frac{\text{HI} - \text{BC}}{\text{AG}}$$

so

$$\text{ED} = \frac{\text{HI} - \text{BC}}{\text{AG}} \cdot \text{AF} + \text{BC}$$

where ofc HI = GI - GH and BC = AC - AB. BUT with your information, you cannot compute AF nor AG, i.e. there is no way to obtain the $x$-position of the middle and rightmost segments, so I say it is impossible.

• You don't need $AF$ or $AG$, you only need $\frac{AF}{AG},$ which you certainly can compute. (We all appear to be assuming the "vertical" lines are all parallel.) – David K Nov 19 '14 at 3:47
• @DavidK Thanks for revealing that implicit assumption. – mvw Nov 19 '14 at 15:48
• I thank everybody for the effort – Nikos Nov 19 '14 at 20:10
• I don't know the values on the x axis The only known is the values on y axis and the vertical lines are parallel – Nikos Nov 19 '14 at 20:13

According to the intercept theorem, it follows that: \begin{align} \frac{ED - BC}{HI - BC}=\frac{FE - AB}{GH - AB} \end{align}

To solve for $ED$, you would get:

\begin{align} ED =\frac{(FE - AB)(HI - BC)}{GH - AB} + BC \end{align}

I hope this helps..

• I believe that this is the solution cause if FD coincide with AC or FD coincide with GI the formula is true Thanks DrC – Nikos Nov 19 '14 at 20:23