The perpendicular from A on side BC of aΔABC intersects BC at D such that DB = 3 CD(see Fig. 6.55).

Applying Pythagoras theorem for ΔACD, we obtain

AC² = AD² + DC²

AD² = AC² – DC²     (i)

Applying Pythagoras theorem in ΔABD, we obtain

AB² = AD² + DB²

AD² = AB² – DB²     (ii)

From (i) and (ii), we get

AC² – DC² = AB² – DB²     (iii)

It is given that 3DC = DB

Therefore,

Putting these values in (iii), we get

16AC² – BC² = 16AB² – 9BC²

16AB² – 16AC² = 8BC²

2AB² = 2AC² + BC²