In Fig. 6.60, AD is a median of a triangle ABC and AM⊥ BC. Prove that.

(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