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Condition for Concurrency of Three Lines

Let ABC and P...

Question

Let $A B C$ and $P Q R$ be any two triangles in the same plane. Assume that the perpendiculars from the points $A, B, C$ to the sides $Q R, R P, P Q$ respectively are concurrent. Using vector methods or otherwise, prove that the perpendiculars from $P, Q, R$ to $B C, C A, A B$ respectively are also concurrent.

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Solution

Let the co-ordinates of the vertices of the $Δ A B C be A (a_{1}, b_{1}), B (a_{2}, b_{2}) and C (a_{3}, b_{3})$ and co-ordinates of the vertices of the $Δ P Q R$ be
$P (x_{1}, y_{1}), B (x_{2}, y_{2}) and R (x_{3} . y_{3})$
Slope of QR $= \frac{y_{2} - y_{3}}{x_{2} - x_{3}}$
$\Rightarrow$ Slope of straight line perpendicular to
$Q R = - \frac{x_{2} - x_{3}}{y_{2} - y_{3}}$
Equation of straight line passing through $A (a_{1}, b_{1},)$ and perpendicular to QR is
$y - b_{1} = - \frac{x_{2} - x_{3}}{y_{2} - y_{3}} (x - a_{1})$ $\Rightarrow (x_{2} - x_{3}) x + (y_{2} - y_{3}) y - a_{1} (x_{2} - x_{3}) - b_{1} (y_{2} - y_{3}) = 0 . . . . (1)$
Similarly equation of straight line from B and perpendicular to RP is
$(x_{3} - x_{1}) x + (y_{3} - y_{1}) y - a_{2} (x_{3} - x_{1}) - b_{2} (y_{3} - y_{1}) = 0 . . . . (2)$
and $e q^{n}$ of straight line from C and perpendicular to PQ is
$(x_{1} - x_{2}) x + (y_{1} - y_{2}) y - a_{3} (x_{1} - x_{2}) - b_{3} (y_{1} - y_{2}) = 0 . . . . (3)$
As staight lines (1),(2) and (3) are given to be concurrent, we should have
$∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{2} - x_{3} & y_{2} - y_{3} & a_{1} (x_{2} - x_{3}) + b_{1} (y_{2} - y_{3}) x_{3} - x_{1} & y_{3} - y_{1} & a_{2} (x_{3} - x_{1}) + b_{2} (y_{3} - y_{1}) x_{1} - x_{2} & y_{1} - y_{2} & a_{3} (x_{1} - x_{2}) + b_{3} (y_{1} - y_{2}) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0 . . . . (4)$
Operating $R_{1} \to R_{1} + R_{2} + R_{3},$ we get
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & S x_{3} - x_{1} & y_{3} + y_{1} & a_{2} (x_{3} - x_{1}) + b_{2} (y_{3} - y_{1}) x_{1} - x_{2} & y_{1} + y_{2} & a_{3} (x_{1} - x_{2}) + b_{3} (y_{1} - y_{2}) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$
Where
$[S = a_{1} (x_{2} - x_{3}) + b_{1} (y_{2} - y_{3}) + a_{2} (x_{3} - x_{1}) + b_{2} (y_{3} - y_{1}) + a_{3} (x_{1} - x_{2}) + b_{3} (y_{1} - y_{2})]$
Expanding along $R_{1}$
$\Rightarrow [(x_{3} - x_{1}) (y_{1} - y_{2}) - (x_{1} - x_{2}) (y_{3} - y_{1})] S = 0$
$\Rightarrow [\frac{y_{1} - y_{2}}{x_{1} - x_{2}} - \frac{y_{3} - y_{1}}{x_{3} - x_{1}}] S = 0 \Rightarrow [m_{P Q} - m_{P R}] S = 0 \Rightarrow S = 0 [m_{P Q} = m_{P R} \Rightarrow P Q ∥ P R;which is not possible in Δ P Q R] \Rightarrow a_{1} (x_{2} - x_{3}) + b_{1} (y_{2} - y_{3}) + a_{2} (x_{3} - x_{1}) + b_{2} (y_{3} - y_{1}) + a_{3} (x_{1} - x_{2}) + b_{3} (y_{1} - y_{2}) = 0.... (5) \Rightarrow x_{1} (a_{3} - a_{2}) + y_{1} (b_{3} - b_{2}) + x_{2} (a_{1} - a_{3}) + y_{2} (b_{1} - b_{3}) + x_{3} (a_{2} - a_{1}) + y_{3} (b_{2} - b_{1}) = 0.... (6)$
(Rearranging the equation (5))
But above condition shows
$∣ ∣ ∣ ∣ ∣ \begin{matrix} a_{3} - a_{2} & b_{3} - b_{2} & x_{1} (a_{3} - a_{2}) + y_{1} (b_{3} - b_{2}) a_{1} - a_{3} & b_{1} - b_{3} & x_{2} (a_{1} - a_{3}) + y_{2} (b_{1} - b_{3}) a_{2} - a_{1} & b_{2} - b_{1} & x_{3} (a_{2} - a_{1}) + y_{3} (b_{2} - b_{1}) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0 . . . (7)$
[Using the fact that as (4) $\Leftrightarrow$ (5) in the same way (6) $\Leftrightarrow$ (7)]
Clearly equaton (7) shows that lines through P and perpendicular to BC, through Q and perpendicular to AB are conurrent.
HENCE PROVED

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