Question

if PL,QM and RN are the altitudes of triangle PQR,whose orthocentre is O,then Q is the orthocentre of

Answer 1

From given information, the solution will be-

PL ⊥ QR, QM ⊥ PR, RN ⊥ PQ
H is intersection of all adove making diffrent triangles in the triangle  PQR
Triangle QLH is a right angled triangle right angled at L then L is the orthocentre of QLH.
Triangle PQL is a right angled triangle right angled at L then L is the orthocentre of PQL.
Triangle QMP is a right angled triangle right angled at M then P is the orthocentre of QMP.
But in any case Q is not the orthocentre of anyone.
Hint- Make triangle and solve as I solved

Answer 2

Given that,

PL, QM and RN are the altitudes of the triangle PQR.

So, PL ⊥ QR, QM ⊥ PR, RN ⊥ PQ

Now, Given Orthocentre of ΔPQR = O

Directly we can say that if a triangle is formed any three points of PQRO, then the fourth one will be Orthocentre.

This way, Q is the Orthocentre of Δ OPR

For the Δ OPR,

OM is already perpendicular to PR, on producing it further we reach Q.

Since POR is obtuse angled, Orthocentre lies outside this triangle.

For line OR, We get a altitude outside the triangle that is, LR

For line OP, We get a altitude outside that is NP

All these meet at Q ( that's how triangle ΔPQR is).

Hence, Q is Orthocentre of ΔPOR.