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.

Answer 2

Let BD = a and DF = b.

In ΔFDC and ΔFBA,

∠FDC = ∠FBA [Both = 90°]

∠DFC = ∠BFA [Common angle]

∴ ∆FDC ~ ∆FBA [By AA]

Since, corresponding sides of similar triangle are proportional to each other.

⇒ `(DF)/(BF)​ = (DC)/(AB)​`

⇒ `(BD+DF)/DF ​= x/z​`

⇒ `b/(a+b) = z/x `     ​.....(1)

In ΔBDC and ΔBFE,

∠BDC = ∠BFE [Both = 90°]

∠DBC = ∠FBE [Common angle]

∴ ∆BDC ~ ∆BFE [By AA]

Since, corresponding sides of similar triangle are proportional to each other.

⇒ `(BD)/(BF​) = (CD)/(EF)​`

⇒ `(BD​)/(BD + DF) = z/y​`

⇒ `a/(a + b) ​= z/y​​`      ​.....(2)

Adding (1) and (2) we get:

⇒ `b/(a + b) ​+ a/(a + b) ​= z/x ​+ z/y​`

⇒ `(a+b)/(a+b​) = z(1/x ​+ 1/y​)`

⇒ `1 = z (1/x ​+ 1/y)`

⇒ `1/z = 1/x ​+ 1/y​`