In ΔABC, D is the midpoint of BC and AE BC. If AC > AB, show that

AB2 = AD2 — BC • DE + 1/4 BC2

In right-angled triangle AED, applying Pythagoras theorem,



AB2 = AE2 + BE2


AE2 = AB2 – BE2 ….(i)


In right-angled triangle AED, applying Pythagoras theorem,


AD2 = AE2 + ED2


AE2 = AD2 – ED2 ….(ii)


Therefore,


AB2 – BE2 = AD2 – ED2





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