Thread: 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'= (AA'+BB'+CC')

Thanks for you help

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'= (AA'+BB'+CC')

Thanks for you help

This is best done using vectors. Let P be a point on the line , and let be the position vector of P. Let be a unit vector perpendicular to . Given a point X, with position vector , let X' be the foot of the perpendicular from X to . Then the position vector of X' is , where d is the distance XX'. Also, the line PX' is perpendicular to , and so (where the dot represents the inner product). Therefore (because ).

If A, B, C, Z have position vectors , then . Thus , which simplifies to .

Thanks Opalg, but actually I was looking for a solution only using Euclidean Geometry. Can you help me?