In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :
(i) AC2 = AD2 + BC.DM + 2 (BC/2)2
(ii) AB2 = AD2 – BC.DM + 2 (BC/2)2
(iii) AC2 + AB2 = 2 AD2 + 1/2BC2
(i) By applying Pythagoras Theorem in ∆AMD, we get,
AM2 + MD2 = AD2 ………………. (i)
Again, by applying Pythagoras Theorem in ∆AMC, we get,
AM2 + MC2 = AC2
AM2 + (MD + DC) 2 = AC2
(AM2 + MD2 ) + DC2 + 2MD.DC = AC2
From equation(i), we get,
AD2 + DC2 + 2MD.DC = AC2
Since, DC=BC/2, thus, we get,
AD2 + (BC/2) 2 + 2MD.(BC/2) 2 = AC2
AD2 + (BC/2) 2 + 2MD × BC = AC2
Hence, proved.
(ii) By applying Pythagoras Theorem in ∆ABM, we get;
AB2 = AM2 + MB2
= (AD2 − DM2) + MB2
= (AD2 − DM2) + (BD − MD) 2
= AD2 − DM2 + BD2 + MD2 − 2BD × MD
= AD2 + BD2 − 2BD × MD
= AD2 + (BC/2)2 – 2(BC/2) MD
= AD2 + (BC/2)2 – BC MD
Hence, proved.
(iii) By applying Pythagoras Theorem in ∆ABM, we get,
AM2 + MB2 = AB2 ………………….… (i)
By applying Pythagoras Theorem in ∆AMC, we get,
AM2 + MC2 = AC2 …………………..… (ii)
Adding both the equations (i) and (ii), we get,
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
2(AM2+ MD2) + (BC/2) 2 + (BC/2) 2 + 2MD (-BC/2 + BC/2) 2 = AB2 + AC2
2AD2 + BC2/2 = AB2 + AC2