Question

The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD . Prove that 2AB² = 2AC² + BC².

Answer 1

Applying Pythagoras theorem for ΔACD, we obtain

AC² = AD² +DC²

AD² = AC² - DC²  ... (1)

Applying Pythagoras theorem in ΔABD, we obtain

AB² = AD² + DB²

AD² = AB² - DB² ...(2)

From equation 1 and 2 we obtain

AC² - DC² = AB² - DB² ....(3)

it is given that 3DC = DB

`:.DC = (BC)/4 `

Putting these value in equation 3 we obtain

`(AC)^2 - ((BC)/4) ^2 = (AB)^2 - ((3BC)/4)^2`

`(AC)^2 - (BC)^2/16 = (AB)^2-`

16AC² - BC² = 16AB² - 9BC²

16AB² - 16AC²  8BC²

2AB² = 2AC² + BC²