Question

Let `ABC and PQR` be any two triangles in the same plane. Assume that the perpendiculars from the points `A, B, C` to the sides `QR, RP, PQ` respectively are concurrent. Using vector methods or otherwise,prove that the perpendiculars from `P, Q, R to BC, CA, AB` respectively are also concurrent.

Answer 1

Answer - Let the position vectors of A,B,C be `vec(a) ,vec(b) vec( c)` and `vec( c)` respectively and that of P,Q,R be `vec(p) , vec(q)` and `vec(r )` respectively . Let `vec(h)` be the the position vectors of the orthocentre H of the `Delta PQR` . We have `HP bot QR.`
Equation of straignt line passing throught A and perpendicular to QR i.e., parallel to HP=`vec(P)-h` is
`vec( r) =vec(a) + l_(1) (vec(p)- vec(h))`
where `t_(1)` is parameter.
Similarly equations of straight line through B and perpendicular to PR is `vec(r ) = vec( b) + t_(2) (vec(q) - vec(h))`
Again equation of straight line through C and perpendicular to `PQ " is " vec(r ) = vec( c) + t_(3) (vec(r ) - hat(h))`
If the lines (i) , (ii) and (iii) are concurrent then there exists a point D with position vector `vec(d)` which lies on all of them that is for some values of `t_(1) , t_(2)` and `t_(3)` which implies that
`(1)/(t_(1)) vec(d) =(1)/(t_(1)) vec(a) + vec(p) - vec(h)`
`(1)/(t_(2))vec(d) =(1)/(t_(2)) vec(b) +vec(q) - vec(h) ......(v)`
`(1)/(t_(3)) vec(d) =(1)/(t_(3)) vec( c) +vec( r) - vec(h)......(vi)`
From Eqs (iv) and (v)
`((1)/(t_(1)) -(1)/(t_(3))) vec(d) =(1)/(t_(1)) vec(a) -(1)/(t_(2)) vec(b) + vec(p) -vec(q)....(vii)`
and from Eqs (v) and (vi)
`((1)/(t_(2)) -(1)/(t_(3))) vec(d) =(1)/(t_(2)) vec(b)- (1)/(t_(3)) vec( c) + vec(q) + vec(r )`
Eliminating `vec(d)` from Eqs (vii) and (viii) we get
`=((1)/(t_(1)) -(1)/(t_(2))) [(1)/(t_(2)) .vec(b) -(1)/(t_(3)) vec(c )+ vec(q) -vec(r )]`
` rArr (t_(3) -t_(2)) [t_(2) vec(a) -t_(1) vec(b) + t_(1)t_(2) (vec(p)-vec(q))]`
` =(t_(2)-t_(1)) [t_(3) vec(b) -l_(2) vec(c) +l_(2)l_(3) (vec(q)-vec(r ))]` [multiplying both sides by `t_(1) t_(2) t_(3)`]
`rArr t_(2)(t_(3)-t_(2))vec(a) +t_(2)(t_(1)-t_(3))vec(b)`
`+t_(2) (t_(2)-t_(1))vec(c) +t_(1)t_(2)(t_(3)-t_(2))vec(p)`
`+t_(2)^(2) (t_(1) -t_(3))vec(q) +t_(2)t_(3)(t_(2) -t_(1))vec(r) =vec(0)`
Thus lines (i) , (ii) and (iii) are concurrent is equivalent to say that there exist scalarst `t_(1),t_(2) " and " t_(3)` such that
`(t_(2)-t_(3))vec(a) +(t_(3)-t_(1))vec(b) +(t_(1)-t_(2))vec(c)+t_(1)(t_(2)-t_(3))vec(p)`
`+t_(2)(t_(3)-t_(1))vec(q)+t_(3)(t_(1)-t_(2))vec(r ) - vec(0)`
On dividing by `t_(1) t_(2) t_(3)` we get
`(lambda_(2)-lambda_(3))vec(p)+(lambda_(3)-lambda_(1)) vec(q) +(lambda_(1)-lambda_(2))vec(r)`
`+lambda_(1)(lambda_(2)-lambda_(3))vec(a) +lambda_(2)(lambda_(3)-lambda_(1))vec(b) +lambda_(3)(lambda_(1) -lambda_(2)) vec(c )=0`
where` lambda_(1) =(1)/(t_(1)) " for " i = 1 ,2,3`
So this is the condition that the lines from p perpendicular to BC from Q perpendicular to CA and from R perpendicular to AB are concurrent (by changing ABC and PQR simultaneously) .