If \[ABC\] is a triangle whose orthocenter is \[P\], circumcenter is \[Q\] then prove that \[\overline {QA} + \overline {QB} + \overline {QC} = \overline {QP} \]

Hint: Here we need to find the following relation between the given vectors. We will first draw the triangle based on the given information. We will consider the LHS of the given equation and simplify it as per the diagram. We will then use the relation between orthocenter, centroid and the circumcenter and form an equation. We will simplify it further and prove \[\overline {QA} + \overline {QB} + \overline {QC} = \overline {QP} \].

Complete step-by-step answer:
First, we will draw the triangle with orthocenter, centroid and the circumcenter. Let \[C'\] be the centroid of the triangle.

Here we need to prove \[\overline {QA} + \overline {QB} + \overline {QC} = \overline {QP} \].
Now, we will consider the left hand side of the given equation.
We can write these vectors as
\[\overline {QA} + \overline {QB} + \overline {QC} = \overline A - \overline Q + \overline B - \overline Q + \overline C - \overline Q \]
On adding and subtracting the like vectors, we get
$\Rightarrow$ \[\overline {QA} + \overline {QB} + \overline {QC} = \overline A + \overline B + \overline C - 3\overline Q \]
Now, multiplying and dividing by 3, we get
$\Rightarrow$ \[\overline {QA} + \overline {QB} + \overline {QC} = 3\left( {\dfrac{{\overline A + \overline B + \overline C }}{3} - \overline Q } \right)\]
We know the relation \[\dfrac{{\overline A + \overline B + \overline C }}{3} = \overline {C'} \] because \[P\] is the orthocenter.
On substituting this value here, we get
$\Rightarrow$ \[\overline {QA} + \overline {QB} + \overline {QC} = 3\left( {\overline {C'} - \overline Q } \right)\]
We can write these vectors as
$\Rightarrow$ \[\overline {QA} + \overline {QB} + \overline {QC} = 3\overline {QC'} \] ……… \[\left( 1 \right)\]
We know that the centroid, orthocenter and circumcenter of a triangle are always collinear and centroid divides the orthocenter and circumcenter in \[1:2\] .





From the diagram we can write
\[\overline {C'} = \dfrac{{2\overline Q + \overline P }}{3}\]
On cross multiplying the terms, we get
\[ \Rightarrow 3\overline {C'} = 2\overline Q + \overline P \]
On further simplification, we get
\[ \Rightarrow 3\overline {C'} - 2\overline Q = \overline P \]
Now, we will subtract \[\overline Q \] from both sides. Therefore, we get
\[ \Rightarrow 3\overline {C'} - 2\overline Q - \overline Q = \overline P - \overline Q \]
On subtracting the like terms, we get
\[ \Rightarrow 3\left( {\overline {C'} - \overline Q } \right) = \overline P - \overline Q \]
On further simplification, we get
\[ \Rightarrow 3\overline {QC'} = \overline {QP} \]
Now, we will substitute this value in equation \[\left( 1 \right)\], therefore, we get
\[ \Rightarrow \overline {QA} + \overline {QB} + \overline {QC} = \overline {QP} \]
Hence, we have proved the following relation between the vectors.

Note: We need to remember that the centroid, orthocenter and circumcenter of a triangle are always collinear i.e. they are on the same straight line. Orthocenter is defined as a point that is formed by the intersection of all three altitudes of the triangle and circumcenter of the triangle is defined as the point which is formed when we draw a circumcircle of the triangle.