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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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