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Question
In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
`"AC"^2 = "AD"^2 + "BC"."DM" + (("BC")/2)^2`
Solution
Applying Pythagoras theorem in ΔAMD, we obtain
AM2 + MD2 = AD2 … (1)
Applying Pythagoras theorem in ΔAMC, we obtain
AM2 + MC2 = AC2
AM2 + (MD + DC)2 = AC2
(AM2 + MD2) + DC2 + 2MD.DC = AC2
AD2 + DC2 + 2MD.DC = AC2 [Using equation (1)]
Using the result, DC = `"BC"/2`, we obtain
`"AD"^2+(("BC")/2)^2 + 2"MD".(("BC")/2) = "AC"^2`
`"AD"^2+(("BC")/2)^2 + "MC" xx "BC" = "AC"^2`
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