Question

# In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that: (i) (ii) (iii)

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Solution

## (i) 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, , we obtain (ii) Applying Pythagoras theorem in ΔABM, we obtain AB2 = AM2 + MB2 = (AD2 − DM2) + MB2 = (AD2 − DM2) + (BD − MD)2 = AD2 − DM2 + BD2 + MD2 − 2BD × MD = AD2 + BD2 − 2BD × MD (iii)Applying Pythagoras theorem in ΔABM, we obtain AM2 + MB2 = AB2 … (1) Applying Pythagoras theorem in ΔAMC, we obtain AM2 + MC2 = AC2 … (2) Adding equations (1) and (2), we obtain 2AM2 + MB2 + MC2 = AB2 + AC2 2AM2 + (BD − DM)2 + (MD + DC)2 = AB2 + AC2 2AM2+BD2 + DM2 − 2BD.DM + MD2 + DC2 + 2MD.DC = AB2 + AC2 2AM2 + 2MD2 + BD2 + DC2 + 2MD (− BD + DC) = AB2 + AC2

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