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


5
1