Sunday, 28 August 2016

Geometric Proof of Parallelogram Law

Parallelogram Law: The sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals.

Though this law is taught and proved using vectors, trigonometry and sometimes co-ordinate geometry, I did not come across any geometric proofs.


So I have written down the below proof using simple geometry for someone who is new to trigonometry or vectors.






In the given Parallelogram ABCD, draw perpendicular at B to CD meeting at F. Similarly draw a perpendicular at C to AB. 
Extend AB to meet the perpendicular at E.  (FC = BE)

AEC, BEC, BFD and BCF form right angled triangles.


From Pythagoreas Theorem,



AE 2  +  CE 2   =  AC 2


AB + BE + 2 AB . BE + C2    =  AC 2            


BE 2  +  CE 2   =  BC 2  



AB  BC 2  + 2 AB . BE    =  AC 2                                                                  ...............(1)

                                                                                                            
                                                                                                                                


DF + BF 2   =  BD 2



CD + FC  2 CD . FC + BF 2    =  BD 2    



FC 2  +  BF 2   =  BC 2



CD + BC 2  2 CD . FC    =  BD 2                                                                    ...............(2)




ADDING (1) AND (2) 



AB + BC + CD + DA 2   =  AC 2  BD 2    


(CD= AB, BC=AD)



  hope this helps!!!




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