Question

In a right triangle ABC,right angled at C,a point D is taken on AB such that CD is perpendicular to AB.Prove that 1/AB^2+1/BC^2=1/CD^2.​

Answer 1

SOLUTION

Ar of ∆ABC= 1/2×Base× Height

=) 1/2× BC× AC= 1/2ab

Ar of ∆ABC= 1/2× Base × Height

=)1/2× AB× CD= 1/2cp

=) 1/2ab= 1/2cp

=) ab= cp..............(1)

Therefore,

In right angled ∆ABC,

=) AB^2 =BC^2 +AC^2

=) (ab/p)^2= a^2 + b^2

=) a^2b^2/p^2= a^2+ b^2....from proof(1)

=) 1/p^2= (a^2+b^2)/a^2b^2

=) 1/p^2= (a^2/a^2b^2+b^2/a^2b^2)

=) 1/p^2= (1/b^2 + 1/a^2)

=) 1/p^2 = (1/a^2+ 1/b^2)

Hence proved

hope it helps ☺️