Question -
Answer -
Given, D and E are points on the sides CA andCB respectively of a triangle ABC right angled at C.
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By Pythagoras theorem in ╬ФACE, we get
AC2 + CE2 =AE2 тАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАж.(i)
In ╬ФBCD, by Pythagoras theorem, we get
BC2 + CD2 =BD2 тАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАж..(ii)
From equations (i) and (ii),we get,
AC2 + CE2 +BC2 + CD2 = AE2 +BD2 тАжтАжтАжтАж..(iii)
In ╬ФCDE, by Pythagoras theorem, we get
DE2 = CD2 +CE2
In ╬ФABC, by Pythagoras theorem, we get
AB2 = AC2 +CB2
Putting the above two values in equation (iii), we get
DE2 + AB2 = AE2 + BD2.