Question

In triangle ABC AD is perpendicular to BC ,BD:CD=3:1 then prove that AB2=AC2 + 1/2 BC2.

Answer 1

given AD perpendicular to BC

AB^2-AC^2=1/2 BC^2

2(AB2-AC2)=BC2

take lhs to prove rhs

2(AB2-AC2)

=2(AD2+BD2-AD2-CD2)

=2{(BD+CD)(BD-CD)}

=2{BC(3CD-CD)

=2(BC)(2CD)

=(BC)(4CD)

=(BC)(3CD+CD)

=(BC)(BC)

=BC2

hope  it helps

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