Let H1 and H2 be the orthocenters of AQD and ABP respectively. Let a=|DÂQ|=|BÂP|. [ADQ]=[ABP] <=> 0,5.|AD|.|AQ|.sin a = 0,5.|AB|.|AP|.sin a <=> |AD|.|AQ|.cos a = |AB|.|AP|.cos a <=> AD.AQ - AB.AP = 0 (this is some vectors) AD.AQ - AB.AP = (AB + BH2 + H2H1 + H1D) (AC + CQ) - AB (AB + BH2 + H2P) = AB.AC + AB.CQ + BH2.AC + BH2.CQ + H2H1.AC + H2H1.CQ + H1D.AC + H1D.CQ - AB.AB - AB.BH2 - AB.H2P = AB.BC + AH1.CQ + BH2.BC + H2H1.AC + H1D.AQ - AB.H2P = BC.AH2 + H2H1.AC = H2H1.AC So [ADQ]=[ABP] <=> H1H2 is perpendicular to AC.

Is any body have a alternate solution ???

Let $H_1,H_2$ be the orthocentres of $\triangle ABP,\triangle ADQ$ respectively. Let $X,Y$ be the feet from $A$ to $CB,CD$ respectively. Let $Z_1,Z_2$ be the feet of the perpendiculars from $H_1,H_2$ to $AC$. Now we observe: $$AZ_1=AH_1 \cos{\angle CAH_1}=BP \cdot \frac{AX}{AC} \cdot \cot{\angle BAP}=\frac{2[ABP]}{AC \cdot \tan{\angle BAP}}$$So using $\angle{BAP}=\angle{ DAQ}$ we get: $$\frac{AZ_1}{AZ_2}=\frac{[ABP]}{[ADQ]}$$Hence as $H_1H_2 \bot AC \Leftrightarrow AZ_1=AZ_2$ we see $H_1H_2 \bot AC \Leftrightarrow [ABP]=[ADQ]$ as desired.

Let $H_1,H_2$ be the orthocentres of $\triangle ABP,\triangle ADQ$ respectively. Let $X,Y$ be the feet from $B$ to $AP$ and $D$ to $AQ$. Call $Z$ the intersection of $AC$ and $H_1H_2$. If we prove that $BX$ and $DY$ intersects on $AC$, we are done. This result follows from radical axis of circle with diameter $AH_1$ and $AH_2$. Now note that $\triangle ABP = \triangle ADQ$ means that $AB \cdot AP = AD \cdot AQ$, since $\angle ABP = \angle ADQ$. This also gives us the spiral similarity of $\triangle ABQ$ and $\triangle APY$. Note that $BH_1$ and $CH_2$ are all lines drawn from the original points of the triangle to its corresponding sides. I do not know how to continue, due to my extensive lack of knowledge in geometry. Anyone care to help?