Question

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

Answer 1

We have DB = 3 CD ... (i)

Since ΔABD is a right triangle (i) right angled at D

AB² – AD² + BD² … (ii)

By ΔACD is a right triangle right angled at D

AC² = AD² + CD² … (iii)

Subtracting equation (iii) from equation (ii),

we got