Question

In the following figure, AB, CD and EF are perpendicular to the straight line BDF.

If AB = x and CD = z unit and EF = y unit, prove that : `1/x + 1/y = 1/z`

Answer 1

In ΔFDC and ΔFBA,

∠FDC = ∠FBA  ...(Since DC || AB)

∠DFC = ∠BFA   ...(Common angle)

ΔFDC ∼ ΔFBA   ...(AA criterion for similarity)

`=> (DC)/(AB) = (DF)/(BF)`

`=> z/x = (DF)/(BF)`  ...(i)

In ΔBDC and ΔBFE,

∠BDC = ∠BFE  ...(Since DC || FE)

∠DBC = ∠FBE  ...(Common angle)

ΔBDC ∼ ΔBFE  ...(AA criterion for similarity)

`=> (BD)/(BF) = (DC)/(EF)`

`=>  (BD)/(BF) = z/y`  ...(ii)

Adding (i) and (ii), we get

`(BD)/(BF) + (DF)/(BF) = z/y + z/x`

`=> 1 = z/y + z/x`

`=> 1/z = 1/x + 1/y`

Hence proved.