In a triangleABC,AD is a median on BC&AE is perpendicular toBC.Prove thtAB.AB=AD.AD-BC.DE+1/4BC.BC

Asked by  | 4th Mar, 2008, 06:01: PM

Expert Answer:

 

 

 

 

                          

Given:-In the adjoining figure D is mid point of side BC of Triangle ABC and AD is the median from A to side BC .and AE is the perpendicular drawn from A to BC.

 

To Prove:-AB2 = AD2 – BC.DE + ¼ BC2   

 

Proof:- AB2 = AE2 + BE2

            AB2 = AD2 – ED2 + (BD –ED)2

            AB2 = AD2 – ED2 + BD2 + ED2 -2*BD*ED

            AB2 = AD2 – 2*BC/2 * DE + (1/2 BC)2

            AB2 = AD2 – BC * DE + 1/4 BC2

 

Answered by  | 1st Jul, 2008, 10:38: PM

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