In the given figure PA,QB and …

In the given figure we have {tex}P A \perp A C{/tex} and {tex}Q B \perp A C{/tex}.
     {tex}\Rightarrow Q B \| P A{/tex}
In  {tex}\triangle PAC{/tex} and {tex}\triangle Q B C{/tex}, we have

 {tex}\angle QCB= \angle PCA{/tex}      ( Common ) 

 {tex}\angle QBC= \angle PAC{/tex}           ( both are 90^o ).
So by AA similarity rule , {tex}\triangle Q B C \sim \triangle P A C{/tex}.
      {tex}\therefore \frac { Q B } { P A } = \frac { B C } { A C }{/tex}
{tex}\Rightarrow \frac { z } { x } = \frac { b } { a + b }{/tex}. .....................................(i) [by the property of similar triangles]
In {tex}\triangle RAC{/tex} , {tex}Q B \| R C{/tex}.
So, {tex}\triangle Q B A \sim \triangle R C A{/tex}.
{tex}\therefore \frac { Q B } { R C } = \frac { A B } { A C }{/tex}
{tex}\Rightarrow \frac { z } { y } = \frac { a } { a + b }{/tex}. .....................................(ii) [by the property of similar triangles]
Form (i) and (ii), we obtain
{tex}\frac { z } { x } + \frac { z } { y }{/tex}{tex}= \left( \frac { b } { a + b } + \frac { a } { a + b } \right) = 1{/tex}
{tex}\Rightarrow \quad \frac { z } { x } + \frac { z } { y } = 1{/tex}
{tex}\Rightarrow \frac { 1 } { x } + \frac { 1 } { y } = \frac { 1 } { z }{/tex}
or  {tex}\frac { 1 } { x } + \frac { 1 } { y } = \frac { 1 } { z }{/tex}.

Hence proved.