The base BC of triangle ABC is bisected at the point $(p,q)$  and the equations to the sides AB and AC are $px+qy=1$  and $qx+py=1$ . The equation of the median through A is 

Equation of the line through AB and AC is
$(px+qy-1)+\lambda (qx+py-1)=0\_\_\_\_\_\left(i\right)$ 
$\therefore$ It passes through (p,q), then
$(p^2+q^2-1)+\lambda (pq+pq-1)=0$ 
$\therefore \lambda =-\frac{(p^2+q^2-1)}{(2pq-1)}$ 
$\therefore$ From equation (i),
 $(px+qy-1)-\frac{(p^2+q^2-1)}{(2pq-1)}(qx+py-1)=0$
or  $(2pq-1)(px+qy-1)=(p^2+q^2-1)(qx+py-1)$