Question

d and e are points on the sides ca and cb respectively of a triangles abc right angled at c . prove that ae²+bd²=ab²+de²​

Answer 1

Answer:

ae^2 = ac^2 + ce^2

bd^2 = dc^2 + bc^2

add these two

ae^2 + bd^2 = ac^2 + ce^2 + dc^2 + bc^2

= (ac^2 + bc^2) + (ce^2 + dc^2)

For right triangles abc and dce

=ab^2 + de^2

Answer 2

hope it's help u........