GEOMETRY
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For details see post on
26-Feb-2015 at 6:00 am.
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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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