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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 ² + BE ² + 2 AB . BE + CE ²    =  AC ²            


BE ²  +  CE ²   =  BC ²  


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


DF ² + BF ²   =  BD ²


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


FC ²  +  BF ²   =  BC ²


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



ADDING (1) AND (2) 


AB ² + BC ² + CD ² + DA ²   =  AC ² +  BD <sup>2    


(CD= AB, BC=AD)</sup>


  hope this helps!!!