Let ABCD a quadrilateral and O the intersection of its diagonals. Denote by M, N the midpoints of AB and CD and let $H_1$ and $H_2$ the orthocenters of triangles OAB and OCD respectively. Prove that AC=BD if and only if H1H2 and MN are parallel.
I observed by sketching a geogebra figure that if I denote by P and Q the midpoints of AD and BC then H1H2 and PQ are always perpendicular (don’t know how to prove it though). This would help proving one of the implications in our problem.
Please help me solve the problem.
I can only prove that $PQ \bot H_1H_2$.
Clearly, the green circle Q(radius = QB) can be drawn passing through B’ and C’. Similarly, the blue circle P (radius = PA) passes through A’ and D’.
Let X and Y be the points of intersection of the two circles. We need to show $YH_2H_1X$ is a straight line.
It should also be clear that $OD’H_2C’$ and $OA’H_1B'$ are circles with $OH_2$ and $OH_1$ as diameters respectively.
Suppose that $YH_2$ is produced to cut the circle $C’H_2D’O$ at R. Then, $\angle ORH_2 = 90^0$. After joining $H_1R$, we get $H_1RO = 90^0$. This means $YH_2RH_1$ is a straight line.
The whole argument can be similarly stated starting from $XH_1$ produced. Instead of getting my hands dirty again, I just conclude that $XH_1RH_2Y$ is a straight line.
Result follows from noting that $X(H_1RH_2)Y$ is the common chord and PQ is the line of centers.