AC is the long diagonal of a parallelogram ABCD. The perpendiculars from C meet the lines AB and AD at P and Q respectively. Show that AC2 = AB·AP + AD·AQ.
Solution
Let the foot of the perpendicular from D to AC be X. Then AXD is similar to AQC, so AQ/AC = AX/AD. Also CXD is similar to APC, so AP/AC = CX/CD. Hence AQ·AD/AC + AP·CD/AC = AX + CX = AC.
© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03