IIT-JEE 1978Question 1
From a point $O$ inside a triangle $ABC$, perpendiculars $OD$, $OE$, $OF$ are drawn to the sides $BC$, $CA$, $AB$ respectively. Prove that the perpendiculars from $A$, $B$, $C$ to the sides $EF$, $FD$, $DE$ are concurrent.
Points and Straight Lines
AnswerProved: the perpendiculars are concurrent
Solution
$A,E,O,F$ lie on a circle with diameter $AO$, so $AE^2-AF^2=OF^2-OE^2$; similarly $BF^2-BD^2=OD^2-OF^2$, $CD^2-CE^2=OE^2-OD^2$. The sum is $0$, which is exactly the condition for the perpendiculars from $A,B,C$ to $EF,FD,DE$ to be concurrent.