Question

In an acute-angled triangle $ABC,$ a point $D$ lies on the segment $BC.$ Let $O_1,O_2$ denote the circumcentres of $\Delta ABD$ and $\Delta ACD,$ respectively. Prove that the line joining the circumcentre of $\Delta ABC$ and the orthocentre of $\Delta O_1{O}_{2}D$ is parallel to $BC.$

Answer 1

Without loss of generality assume that $\angle ADC\ge {90}^o$.
Let O denote the circumcenter of triangle ABC and K the orthocentre of triangle $O_1{O}_{2}D$.
Note that circumcircles of triangles ABD and ACD pass through the points A and D, so AD is perpendicular to $O_1{O}_2$ and, triangle $A{O}_1{O}_2$ is congruent to triangle $D{O}_1{O}_2$.
$\angle A{O}_1{O}_2=\angle O_2{O}_{1}D=\angle B$ since $O_2{O}_1$ is the perpendicular bisector of AD.
Since $O{O}_2$ is the perpendicular bisector of AC, it follows that $\angle AO{O}_2=\angle B$.
$\therefore O$ lies on the circumcircle of triangle $A{O}_1{O}_2$.
Since AD is perpendicular to $O_1{O}_2$, we have $\angle O_{2}KA={90}^o-\angle O_1{O}_{2}K=\angle O_2{O}_{1}D=\angle B$.
$\therefore K$ also lies on the circumcircle of triangle $A{O}_1{O}_2$.
$\therefore \angle AKO={180}^o-\angle A{O}_{2}O=\angle ADC$ and hence OK is parallel to BC.
Remark. The result is true even for an obtuse-angled triangle.