Hmmm... probably a coordinates approach would work here... But it would be darn messy... I'm now working on it.

We have $H_2M \perp DN$ and $NC \perp DN$ so $H_2M \parallel NC$. Similarly, we have $H_2N \parallel MC$. It follows that $H_2MCN$ is a parallelogram. $H_2C \cap MN=T$ then $T$ is the midpoint of $MN$ and $H_2C$. We have $MH_1 \parallel DN,NH_1 \parallel DM$ so $DMH_1N$ is a parallelogram. It follows that $T$ is also midpoint of $DH_1$. Thus, $H_1H_1CD$ is a parallelogram which means $H_1H_1=DC$ and $H_2H_1 \parallel DC$ or $H_2H_1 \perp AB$. Thus, $[H_1AH_1B]= \frac{H_2H_1 \cdot AB}{2}= \frac{CD \cdot AB}{2}=S$. $\blacksquare$

Oh my g**!!! I'm so frustrated... I got those 2 parallelograms in a jiffy and then got stuck... At least - new technique added to toolchest - concentric parallelograms! Anyone wanna try coordinates?

Setting the origin of the plane in the center $O$ of the circle $(NDMC)$ and denoting $x=\overrightarrow{OX}$, we have that $h_1=m+n+c, h_2=n+d+m$, hence $\overrightarrow{H_1H_2}=h_2-h_1=d-c=\overrightarrow{CD}$, which kind of trivializes the problem..

shinichiman wrote: We have $H_2M \perp DN$ and $NC \perp DN$ so $H_2M \parallel NC$. Similarly, we have $H_2N \parallel MC$. It follows that $H_2MCN$ is a parallelogram. $H_2C \cap MN=T$ then $T$ is the midpoint of $MN$ and $H_2C$. We have $MH_1 \parallel DN,NH_1 \parallel DM$ so $DMH_1N$ is a parallelogram. It follows that $T$ is also midpoint of $DH_1$. Thus, $H_1H_1CD$ is a parallelogram which means $H_1H_1=DC$ and $H_2H_1 \parallel DC$ or $H_2H_1 \perp AB$. Thus, $[H_1AH_1B]= \frac{H_2H_1 \cdot AB}{2}= \frac{CD \cdot AB}{2}=S$. $\blacksquare$ A bit of typos here: Thus, $H_2H_1CD$ is a parallelogram which means $H_2H_1=DC$ and $H_2H_1 \parallel DC$ or $H_2H_1 \perp AB$. Thus, $[H_2AH_1B]= \frac{H_2H_1 \cdot AB}{2}= \frac{CD \cdot AB}{2}=S$. $\blacksquare$

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2587358&sid=7f6151a18569bd348857a2a9040be1ad#p2587358

Its easy to prove $CH_1\parallel DH_2$. And $DH_2=2Rcos\angle NDM=2Rcos\angle CNM=CH_1$ where $R$ - radius of circumcircle $DNCM$. So $H_1CDH_2$ a parallelogram. $2A( H_1CDH_2)= H_1H_2 \cdot AB=CD \cdot AB=2S$. And $A( H_1CDH_2)=S$

Why is that in my diagram AH1BH2 is inside triangle ABC? Pls show a figure. Tnx

Solution: Claim: $H_1MDN$ and $H_2MCN$ are parallelograms. Proof: Clearly $CMDN$ is cyclic. So, $\angle MNC = \angle MDC = 90-\angle DCM=A$. $\angle H_1MN=90-A=\angle MCD=\angle MND$. So, $MH_1||ND$. Same way we can get $H_1N||DM$, which gives us $H_1MDN$ is a parallelogram. Same way we can get $H_2MCN$ is parallelogram. Claim: $CDH_2H_1$ is a parallelogram. Proof: Let $T$ be the midpoint of $MN$. As $H_1MDN$ and $H_2MCN$ are parallelograms, $DT=TH_1$ and $CT=TH_2$. So, $CDH_2H_1$ is a parallelogram. Claim: $[AH_1BH_2]=S$. Proof: As $CDH_2H_1$ is a parallelogram, $H_1H_2||CD \perp AB$ and $CD=H_1H_2$. So, \[S=\frac{1}{2}\cdot AB\cdot CD=\frac{1}{2}\cdot AB \cdot H_1H_2 = [AH_1B]+[AH_2B]=[AH_1BH_2].\] Proved. $\blacksquare$

I just immediately showed that CH1H2D is a parallelogram by showing that CH1 is parallel to DH2 (both perpendicualr to MN) and CH1=DH2 (both are the length MN times cotC since a quick angle chase shows that angles MH2 N and ACB are equal). I didn't any other parallelograms in the figure. Also ignore my question last time I was stupid last year.

Let $N'=DN\cap MH_2$ and by construction, $\angle NN'M=90^{\circ}.$ Let $M'=MH_1\cap BC$. Since $DN\parallel MH_1$ and $\angle N'NM'=\angle MM'N=90^{\circ}$, $N'MM'N$ is a rectangle. Hence, $H_2M\parallel BC$. Note that $NH_1\perp AC$ and $DM\perp AC$ so $DM\parallel NH_1$. Also, note that $H_2D\perp MN$ and $CH_1\perp MN$ so $H_2D\parallel H_1C$. Because $DMH_1N$ is a parallelogram, $NH_1=DM$ and by ASA, $\triangle H_2DM \cong \triangle CH_1N \implies CH_1=DH_2$. Since we also have $CH_1\parallel H_2D$, these prove that $H_2DCH_1$ is a parallelogram so $H_1H_2\parallel CD\implies H_1H_2\perp AD$ and $H_2H_1=CD$. Thus, $$[AH_1BH_2]=\frac{1}{2}\cdot AB \cdot H_1H_2=\frac{1}{2}\cdot AB\cdot CD=[ABC]=S$$as desired. $\blacksquare$

Put $(MCND)$ on a unit circle in complex plane. This gives us $H_1 = m+n+c$ and $H_2=m+n+d$ $\implies H_1H_2=c-d=CD$ and $H_1C=m+n=H_2D$ $\implies CDH_1H_2 $ is a parallelogram. $\implies H_1H_2 \perp AB$ let $H_1H_2 \cap AB = X$ and let $H_1X=\omega , H_2X=CD - \omega$. So we get $[AH_1BH_2] = \frac{AB \cdot CD}{2}=S$