Question

Triangle ABC is right angled at B and D is the midpoint of BC. prove that AC=4AD_3AB

Answer 1

I think question will be this

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

Answer 2

AD^2= AB^2 + BD^2 (PYTH THEO )

=AB^2 + BC^2/4

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

4AD^2 - 4AB^2 = BC^2 - 1 EQUATION

AC^2 = AB^2+ BC ^2 (PYTH THEO) - 2 EQUATION

SUBSITUTUING THE VALUE OF 1 EQUATION IN 2 EQUATION

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

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