Answers

2014-07-22T22:47:32+05:30
In ΔABC angle c =90
AB^2=AC^2+BC^2
CQ=BC/2,  CP=AC/2
from Pythagoras theorem,
   AQ^2=AC^2+CQ^2
   BP^2=PC^2+BC^2
L.H.S=4(AQ^2+BP^2)
          =4(AC^2+CQ^2+PC^2+BC^2)
          =4(AC^2+BC^2)+4(CQ^2+PC^2)
          =4AB^2+4{(BC/2)^2+(AC/2)^2}             (BECAUSE P & Q ARE MID POINT OF AC & BC)
           =4AB^2+4(BC^2+AC^2)/4
            = 4AB^2+AB^2
   L.H.S=5AB^2
   L.H.S=R.H.S
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