# Perpendiculars

• Feb 10th 2011, 08:36 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'= $\frac{1}{3}$(AA'+BB'+CC')

Thanks for you help
• Feb 13th 2011, 11:34 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'= $\frac{1}{3}$(AA'+BB'+CC')

Thanks for you help

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

If A, B, C, Z have position vectors $\mathbf{a},\ \mathbf{b},\ \mathbf{c},\ \mathbf{z}$, then $\mathbf{z} = \frac13(\mathbf{a}+\mathbf{b}+ \mathbf{c})$. Thus $\frac13(AA'+BB'+CC') = \frac13\bigl((\mathbf{a} - \mathbf{p})\mathbf{.}\mathbf{n} + (\mathbf{b} - \mathbf{p})\mathbf{.}\mathbf{n} + (\mathbf{c} - \mathbf{p})\mathbf{.}\mathbf{n}\bigr)$, which simplifies to $(\mathbf{z} - \mathbf{p})\mathbf{.}\mathbf{n} = ZZ'$.
• Feb 13th 2011, 11:46 AM
bram kierkels
Thanks Opalg, but actually I was looking for a solution only using Euclidean Geometry. Can you help me?