From the given figure, find the relation between x, y and z if AB, EF and DC are perpendicular to BC.

$xy=z(y - x)$

In ΔABC,

BF = a, FC= b

EF is parallel to AB

\(\frac{EF}{AB}\) = \(\frac{FC}{BC}\) 

= \(\frac{x}{z}\) = \(\frac{b}{a + b}\)  ....(i)

In ΔBDC,

\(\frac{EF}{DC}\) = \(\frac{BF}{BC}\)

= \(\frac{x}{y}\) = \(\frac{a}{a + b}\) ....(ii)

Add (i)  & (ii),

= \(\frac{x}{z}\) + \(\frac{x}{y}\)  = \(\frac{b}{a + b}\) + \(\frac{a}{a + b}\) = 1

= \(\frac{x}{z}\) + \(\frac{x}{y}\)  = 1

= \(\frac{1}{z}\) + \(\frac{1}{y}\) = \(\frac{1}{x}\)

= \(\frac{1}{x}\) - \(\frac{1}{y}\) = \(\frac{1}{z}\) 

= \(\frac{y - x}{xy}\) = \(\frac{1}{z}\)

= xy = z(y - x)