Let the position vectors of points A,B,C,D be a,b,c and d, respectively.
Then, AB=b−a,BC=c−b,AD=d−a, BD=d−b,CA=a−c,CD=d−c
Now, ∣AB×CD+BC×AD+CA×BD∣ =∣(b−a)×(d−c)+(c−b)×(d−a)+(a−c)×(d−b)∣ =∣b×d−a×d−b×c+a×c+c×d−c×a−b×d +b×a+a×d−a×b−c×d+c×b∣ =2∣a×b+b×c+c×a).....(i)
Also, area of △ABC =21∣AB×AC∣=21∣(b−a)×(c−a)∣ =21∣b×c−b×a−a×c+a×a∣ =21∣a×a+b×c+c×a∣.......(ii)
From Eqs. (i) and (ii), ∣AB×CD+BC×AD+CA×BD∣2(2 area of △ABC) =4( area of ΔABC)