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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`
Solution
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.
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