From the vertices A, B, C of a triangle ABC, perpendiculars AD, BE, CF are drawn to any straight line. Show that the perpendiculars from D, E, F to BC, CA, AB respectively are concurrent.
Text Solution
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Without loss of generality, we can take the line as the x-axis and A, B, C as (x 1 , y 1 ), (x 2 , y 2 ) and (x 3 , y 3 ).
The points D, E, F are (x 1 , 0), (x 2 , 0) and (x 3 , 0).
Equation of the perpendicular from D to BC is
(x – x 1 )(x 3 – x 2 ) + y(y 3 – y 2 ) = 0 ……..(L 1 )
also, the other two perpendiculars are
(x – x 2 )(x 1 – x 3 ) + y(y 1 – y 3 ) = 0 ……..(L 2 )
and (x – x 3 )(x 2 – x 1 ) + y(y 2 – y 1 ) = 0 ……..(L 3 )
We have L 1 + L 2 + L 3 = 0 which implies that the three perpendiculars are concurrent.
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