# Perpendiculars

• Feb 10th 2011, 07:36 AM
bramkierkels
Perpendiculars
(See Attachment)
Given a Triangle ABC with circumcenter Z.
And a line l outside the triangle ABC.
There are four perpendiculars to L, starting in A, B, C and Z. (ending in respectively A',B',C' and Z')
I want to proof:

ZZ'=$\displaystyle \frac{1}{3}$(AA'+BB'+CC')

Thanks for you help
• Feb 13th 2011, 10:34 AM
Opalg
Quote:

Originally Posted by bramkierkels
(See Attachment)
Given a Triangle ABC with circumcenter Z.
And a line l outside the triangle ABC.
There are four perpendiculars to L, starting in A, B, C and Z. (ending in respectively A',B',C' and Z')
I want to proof:

ZZ'=$\displaystyle \frac{1}{3}$(AA'+BB'+CC')

Thanks for you help

This is best done using vectors. Let P be a point on the line $\displaystyle \ell$, and let $\displaystyle \mathbf{p}$ be the position vector of P. Let $\displaystyle \mathbf{n}$ be a unit vector perpendicular to $\displaystyle \ell$. Given a point X, with position vector $\displaystyle \mathbf{x}$, let X' be the foot of the perpendicular from X to $\displaystyle \ell$. Then the position vector of X' is $\displaystyle \mathbf{x} + d\mathbf{n}$, where d is the distance XX'. Also, the line PX' is perpendicular to $\displaystyle \mathbf{n}$, and so $\displaystyle (\mathbf{x}+d\mathbf{n} - \mathbf{p})\mathbf{.}\mathbf{n} = 0$ (where the dot represents the inner product). Therefore $\displaystyle d = (\mathbf{x} - \mathbf{np})\mathbf{.}\mathbf{n}$ (because $\displaystyle \mathbf{n}\mathbf{.}\mathbf{n} = 1$).

If A, B, C, Z have position vectors $\displaystyle \mathbf{a},\ \mathbf{b},\ \mathbf{c},\ \mathbf{z}$, then $\displaystyle \mathbf{z} = \frac13(\mathbf{a}+\mathbf{b}+ \mathbf{c})$. Thus $\displaystyle \frac13(AA'+BB'+CC') = \frac13\bigl((\mathbf{a} - \mathbf{p})\mathbf{.}\mathbf{n} + (\mathbf{b} - \mathbf{p})\mathbf{.}\mathbf{n} + (\mathbf{c} - \mathbf{p})\mathbf{.}\mathbf{n}\bigr)$, which simplifies to $\displaystyle (\mathbf{z} - \mathbf{p})\mathbf{.}\mathbf{n} = ZZ'$.
• Feb 13th 2011, 10:46 AM
bram kierkels
Thanks Opalg, but actually I was looking for a solution only using Euclidean Geometry. Can you help me?