Question

in fig triangle ABC is right angled at B and D is midpoint of BC prove that AC square=4AD square-3AB square

Answer 1

By using pythagoras theorem

Answer 2

In ∆ ABC :-

=> AC² = AB² + BC² -------- (1)

and, in ∆ ABD :-

=> AD² = AB² + BD²
=> AB² = AD² - BD² --------- (2)
=> BD² = AD² - AB² --------- (3)
Taking equation - (1) :-

=> AC² = AB² + BC²
=> AC² = AD² - BD² + BC² --{from - (2)}
=> AC² = AD² - BD² + (BD + DC)²
=> AC² = AD² - BD² + BD² + DC² + 2BD.DC
=> AC² = AD² + DC² + 2BD.DC
=> AC² = AD² + DC² + 2BD.BD
=> AC² = AD² + BD² + 2BD²
=> AC² = AD² + 3BD²
=> AC² = AD² + 3(AD² - AB²) ----{from - (3)}
=> AC² = AD² + 3AD² - 3AB²
=> AC² = 4AD² - 3AB²
Q.E.D.

Hope this helps.....