In the adjoining figure , AB||CD||EF.
prove that `1/x+1/y=1/z`.

Explanation

In `triangleBFE and triangle BDC`,
` angle1 =angle2` ( corresponding `angles ` as EF||CD)
`angle = angle3` (common)

`(BF)/(BD)=(FE)/(DC)` ( corresponding sides of similar triangles are proportional)
`(BF)/(BD)=z/y .....(i)`
In `triangleDFE and triangleDBA`
and `angle4= angle5` ( correspending `angles` as EF||AB)
`angle6= angle6` (common)

`(DF)/(DB) = (FE)/(BA)` (corresponding sides of similar triangles are proportional)
`(DF)/(DB)= z/x ...(2)`
( Remember , common part in (1) and (2) should be in denominator so that we can add easily) Adding (1) and (2) , we get,
`(BF)/(BD)+(DF)/(DB)= z/y+z/x`
`(BF+DF)/(BD)=(1/y+1/x)`
`(BD)/(BD)=z(1/x+1/y)Rightarrow1=z(1/x+1/y)`
`1/z=1/x+1/y`