Question

in triangle ABC angle ABC =90 D is the midpoint of BC then prove that AC2=4AD2-3AB2​

Answer 1

1) in traingle ABC,

angle ABC = 90

By Pythagoras thm,

AC^2 = AB^2 + BC^2 .....(1)

2) in triangle ABD,

angle ABD = 90

By Pythagoras thm,

AD^2 = AB^2 + BD^2

AD^2 = AB^2  + (BC/2)^2  ......(Since D is midpt of BC)

AD^2 = AB^2 + BC^2/4

Multiplying throughout by 4,

4AD^2 = 4AB^2 + BC^2

4AD^2 = 3AB^2 + AB^2 + BC^2

4AD^2 = 3AB^2 + AC^2  .....(From (1))

AC^2 = 4AD^2 - 3AB^2

Hence proved