Let $O$ be the circumcenter. I have written some facts if you don't understand something just ask: Do you understand: $OKH_1N$ is paralelogram. Do you understand: $H_1H_2LN$ is paralelogram.$=>H_1H_2 || LN$. By Analogy we will get $H_3H_4 || LN=>H_1H_2 || H_3H_4$. By Analogy we will get $H_2H_3 || H_1H_4$. Thus $H_1H_2H_3H_4$ is paraleleogram.

we can use complex numbers very easy, but first consider $M$, the symmetric of $A$ w.r.t. $H_1$. M is the ortochentre of $\triangle{ABD}$. Let O be the origin, (O is the center of the circle of $ABCD$), we have $m=a+b+d, h_1=a+\frac{b+d}{2}$, and the others $h_2=b+\frac{a+c}{2}, h_3=c+\frac{b+d}{2},h_4=d+\frac{a+c}{2}$, and all we have to prove is that $h_1+h_3=h_2+h_4$, which is obvious.

MariusBocanu wrote: $m=a+b+d, h_1=a+\frac{b+d}{2}$, and the others $h_2=b+\frac{a+c}{2}, h_3=c+\frac{b+d}{2},h_4=d+\frac{a+c}{2}$, and all we have to prove is that $h_1+h_3=h_2+h_4$, which is obvious. But aren't these are the formulae for the centroids, not the orthocentres? And your final formula is necessay for a parallelogram, but is it also sufficient? Merlin

Merlinaeus wrote: MariusBocanu wrote: $m=a+b+d, h_1=a+\frac{b+d}{2}$, and the others $h_2=b+\frac{a+c}{2}, h_3=c+\frac{b+d}{2},h_4=d+\frac{a+c}{2}$, and all we have to prove is that $h_1+h_3=h_2+h_4$, which is obvious. But aren't these are the formulae for the centroids, not the orthocentres? And your final formula is necessay for a parallelogram, but is it also sufficient? Merlin For the first question, if O, the center of the circumcircle is the origin, $g=\frac{a+b+c}{3},h=a+b+c$, where $a,b,c$ are afixes of the vertices, $g,h$ the afixes of centroid and ortochentre. (just use Euler line) For the second one, a convex quadrilateral ABCD is parallelogram if and only if $a+c=b+d$, which means that the diagonals AC and BD have the same midpoint, which is obvious a necessary and a suficient condition.

Nice problem. Reflect $A,B,C,D$ onto $NK, KL,LM,MN$ respectively, note that these points after reflection converge to the same point $O$(say). Let the orthocentres of $NKO,KLO,LMO$ and $MNO$ be $H_1',H_2',H_3'$ and $H_4'$ respectively. We know that $KLMN$ is a paralellogram. Also note that it is equivalent to prove that $H_1'H_2'H_3'H_4'$ is a parallelogram. So, the problem boils down to proving this problem: Let $KLMN$ be a parallelogram. Suppose $O$ is a point inside it such that $\angle NOK+\angle MOL=180^{\circ}$. Then $H_1'H_2'H_3'H_4'$(defined as above) is a paralellogram. Proof: It is easy to see that $H_1'H_3'\perp ML$. Suppose the circumradius of $ONK$ and $OML$ be $R_1$ and $R_2$ respectively. Easy enough by Sine Law $R_1=R_2$. So, $|OH_1'|=2R_1\cos\angle NOK=2R_2\cos\angle MOK=|OH_3'|$. Thus $H_1'H_2'H_3'H_4'$ is a paralellogram with centre $O$. This finishes our proof.

We consider the circle of ABCD as the unit circle $$H_1=\frac{b+d}{2}+a$$$$H_2=\frac{a+c}{2}+b$$$$H_3=\frac{b+d}{2}+c$$$$H_4=\frac{a+c}{2}+d$$ Let us note that: $$H_1+H_3=H_2+H_4$$QED