pls answer

In EDC and ABC

EDC = ABC  (given)

ECD  is common.

By AA corollary

EDC ABC

⇒ED/ AB = CD/ BC ..............(1)

Similarly, EDB BFC

⇒ED/ FC = BD/BC ...............(2)

Adding (1) and (2)

ED/AB + ED/ FC = CD/BC + BD/BC

ED (1/AB + 1/ FC) = 1/BC (CD + BD)

Now on sunbstituting AB=x,ED=y,FC=z

y (1/x + 1/z) = 1/BC (BC)

y (1/x + 1/z) = 1

1/x + 1/z = 1/y

This is the required proof.