In Fig. 6.60, AD is a median of a triangle ABC and AM⊥ BC. Prove that:
(i) 
(ii) 
(iii) 
(i) Using, Pythagoras theorem in ΔAMD, we get
AM2 + MD2 = AD2 (i)
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 (i)]
Using, DC = BC/2 we get
AD2 + (BC/2)2 + 2MD * (BC/2) = AC2
AD2 + (BC/2)2 + MD * BC = AC2
(ii) Using 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
= AD2 + (BC/2)2 – 2 (BC/2) * MD
= AD2 + (BC/2)2 – BC * MD
(iii)Using 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
2 (AM2 + MD2) + (BC/2)2 + (BC/2)2 + 2MD (-BC/2 + BC/2) = AB2 + AC2
2AD2 + BC2/2 = AB2 + AC2