Solution:
In \[\Delta \text{ }FDC\text{ }and\text{ }\Delta \text{ }FBA,\]
\[\angle FDC\text{ }=\angle FBA\text{ }\left[ As\text{ }DC\text{ }||\text{ }AB \right]\]
\[\angle DFC\text{ }=\angle BFA\] [common angle]
Hence, \[\vartriangle FDC\text{ }\sim\text{ }\vartriangle FBA\text{ }by\text{ }AA\]criterion for similarity.
So, we have
\[DC/AB\text{ }=\text{ }DF/BF\]
\[z/x\text{ }=\text{ }DF/BF\text{ }\ldots .\text{ }\left( 1 \right)\]
In \[\Delta \text{ }BDC\text{ }and\text{ }\Delta \text{ }BFE,\]
\[\angle BDC\text{ }=\angle BFE\text{ }\left[ As\text{ }DC\text{ }||\text{ }FE \right]\]
\[\angle DBC\text{ }=\angle FBE\][Common angle]
Hence, \[\vartriangle BDC\text{ }\sim\text{ }\vartriangle BFE\text{ }by\text{ }AA\]criterion for similarity.
So, we have
\[BD/BF\text{ }=\text{ }z/y\text{ }\ldots ..\text{ }\left( 2 \right)\]
Now, adding (1) and (2), we get
\[BD/BF\text{ }+\text{ }DF/BF\text{ }=\text{ }z/y\text{ }+\text{ }z/x\]
\[1\text{ }=\text{ }z/y\text{ }+\text{ }z/x\]
Thus,
\[1/z\text{ }=\text{ }1/x\text{ }+\text{ }1/y\]
– Hence Proved