Question 37352: If M and N are the midpoints of sides AB and BC of Triangle ABC right angled at B, then prove that 4(AN^2+CM^2)=5AC^2
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! If M and N are the midpoints of sides AB and BC of Triangle ABC right angled at B, then prove that 4(AN^2+CM^2)=5AC^2
ABN IS A RIGHT ANGLED TRIANGLE WITH ANGLE ABN=90.HENCE USING PYTHOGARUS THEOREM
AN^2=AB^2+BN^2....................I
MBC IS A RIGHT ANGLED TRIANGLE WITH ANGLE MBC=90.HENCE USING PYTHOGARUS THEOREM
CM^2=MB^2+BC^2....................II
4*(EQN.I + EQN.II)......GIVES
4(AN^2+CM^2)=4(AB^2+BC^2)+4(BN^2+MB^2).............................III
N IS MID POINT OF BC ,HENCE BN=NC=BC/2
M IS MID POINT OF AB ,HENCE MB=BA=AB/2...SUBSTITUTING IN EQN.III
4(AN^2+CM^2)=4(AB^2+BC^2)+4{(BC/2)^2+(AB/2)^2}
=4(AB^2+BC^2)+AB^2+BC^2........................................IV
BUT ABC IS RIGHT ANGLED TRIANGLE WITH ANGLE ABC=90,WE HAVE
AB^2+BC^2=AC^2...HENCE
4(AN^2+CM^2)=4AC^2+AC^2=5AC^2
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