DB = 3CD (i) [Given]
In ∆ADB, ∠ADB = 90° [Given]
∴ AB2 = AD2 + DB2 [Pythagoras theorem]
∴ AB2 = AD2 + (3CD)2 [From (i)]
∴ AB2 = AD2 + 9CD2 (ii)
In ∆ADC, ∠ADC = 90° [Given]
∴ AC2 = AD2 + CD2 [Pythagoras theorem]
∴ AD2 = AC2 – CD2 (iii)
AB2 = AC2 – CD2 + 9CD2 [From (ii) and(iii)]
∴ AB2 = AC2 + 8CD2 (iv)
CD + DB = BC [C – D – B]
∴ CD + 3CD = BC [From (i)]
∴ 4CD = BC
∴ CD = BC/4 (v)
AB2 = AC2 + 8(BC/4)2 [From (iv) and (v)]
∴ AB2 = AC2 + 8 × BC2/16
∴ AB2 = AC2 + BC2/2
∴ 2AB2 = 2AC2 + BC2 [Multiplying both sides by 2]