Question

(2) In obtuse angled A ABC, ZB > 90°, if seg
AD I side BC and D-B-C, then prove
AC2 = AB + BC2 + 2BC.DB.
A
D
B
C С​

Answer 1

Answer:

Given: ABC is obtuse triangle.

AB² = AD² + DB² ⟶(1)

AC² =AD² + DC² ⟶(2)

from (1) AD² = AB² − DB²

∴AC² = AB² − DB² + (DB+BC)² [∵BC=DB+BC]

AC² = AB² − DB² + DB² + BC ² + 2BC × DB

AC² = AB² + BC² + 2BC × DB

Hence Proved