Question

if anyone can do this I will mark him as brainliest​

Answer 1

Answer:

First, triangles OCB and PCA are similar to each other by AA similarity

(<OCB = <PCA ,  <OBC = <PAC = 90 Degrees)

Therefore, by cpst,

BC/AC = OB/AP

b/(a+b) = z/x ---------------> Eqn. 1

Similarly in trianles OAB and RAC,

by cpst

OB/RC = AB/AC

a/(a+b)=z/y -----------------> Eqn. 2

Eqn. 1 +Eqn. 2 = b/(a+b) + a/(a+b) = (a+b)/(a+b) = 1

                       = z/x +z/y

Therefore, z/x + z/y = 1

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

Finally, 1/x + 1/y = 1/z