In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

1)`AC^2 = AD^2 + BC . DM + ((BC)/2)^2`

2)` AB^2 = AD^2 – BC . DM + ((BC)/2)^2`

3)` AC^2 + AB^2 = 2 AD^2 + 1/2(BC)^2`

#### Solution

(i) Applying Pythagoras theorem in ΔAMD, we obtain

AM^{2} + MD^{2} = AD^{2 }… (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM^{2} + MC^{2} = AC^{2}

AM^{2} + (MD + DC)^{2} = AC^{2}

(AM^{2} + MD^{2}) + DC^{2} + 2MD.DC = AC^{2}

AD^{2} + DC^{2} + 2MD.DC = AC^{2} [Using equation (1)]

Using the result, DC = BC/2, we obtain

`AD^2+((BC)/2)^2 + 2MD.((BC)/2)= AC^2`

`AD^2+((BC)/2)^2 + MC xx BC = AC^2`

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AB^{2} = AM^{2} + MB^{2}

= (AD^{2} − DM^{2}) + MB^{2}

= (AD^{2} − DM^{2}) + (BD − MD)^{2}

= AD^{2} − DM^{2} + BD^{2} + MD^{2} − 2BD × MD

= AD^{2} + BD^{2} − 2BD × MD

`= AD^2+((BC)/2)^2 - 2((BC)/2) xx MD`

`= AD^2 + (`

(iii)Applying Pythagoras theorem in ΔABM, we obtain

AM^{2} + MB^{2} = AB^{2} … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM^{2} + MC^{2} = AC^{2} … (2)

Adding equations (1) and (2), we obtain

2AM^{2} + MB^{2} + MC^{2} = AB^{2} + AC^{2}

2AM^{2} + (BD − DM)^{2} + (MD + DC)^{2} = AB^{2} + AC^{2}

2AM^{2}+BD^{2} + DM^{2} − 2BD.DM + MD^{2} + DC^{2} + 2MD.DC = AB^{2 }+ AC^{2}

2AM^{2} + 2MD^{2} + BD^{2} + DC^{2} + 2MD (− BD + DC) = AB^{2} + AC^{2}

`2(AM^2 + MD^2) + ((BC)/2)^2 + ((BC)/2)^2 + 2MD ((-BC)/2 + (BC)/2) = AB^2+AC^2`

`2AD^2 + (BC^2)/2 = AB^2 + AC^2`