Question

The perpendicular from A on side BC of a Δ ABC meets 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 equation (2), we obtain

AC² - DC² = AB² - DB² .....(3)

It is given that 3DC = DB

∴ `"DC" = "BC"/4` and `"DB"=(3"BC")/ 4`

Putting these values in equation (3), we obtain

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

`"AC"^2 - "BC"^2/16 = "AB"^2 - (9"BC"^2)/4`

16AC² - BC² = 16AB² - 9BC²

16AB² - 16AC² = 8BC²

2AB² = 2AC² + BC²