Question

In the following figure, ∠ABD = ∠CDB = ∠EFD = 90°. If AB = x, CD = y, EF = z, prove that `1/z = 1/x + 1/y`:

Answer 1

Here,

AB = x, CD = y, EF = z, and ∠ABD = ∠CDB = ∠EFD = 90°

From the figure, AB and CD are perpendicular to BD, and lines AD and BC intersect at E. Also EF ⊥ BD.

⇒ In triangles ABE and DFE,

• ∠ABE = 90° and ∠DFE = 90°
• ∠AEB = ∠DEF (vertically opposite angles)

Therefore, △ABE ∼ △DFE, by AA similarity.

So, `(AB)/(DF) = (BE)/(FE) = (AE)/(DE)`

Hence, `(AB)/(EF) = (BE)/(DF) or x/z = (BE)/(DF)`

⇒ In triangles CDE and BFE,

• ∠CDE = 90° and ∠BFE = 90°
• ∠CED = ∠BEF (vertically opposite angles)

Therefore, △CDE ∼ △BFE by AA similarity.

So, `(CD)/(BF) = (DE)/(FE) = (CE)/(BE)`

Hence, `(CD)/(EF) = (DE)/(BF) or Y/z = (DE)/(BF)`

From the similarity of △ABE and △DFE,

`(BE)/(FE) = (DF)/(AB)`

`BE = (z xx DF)/y`

From the similarity of △CDE and △BFE,

`(DE)/(FE) = (BF)/(CD)`

`DE = (z xx BF)/y`

A simpler route is to use the corresponding proportional results as:

`(DF)/x = (BE)/x, (BF)/z = (DE)/y`

But, BF + FD = BD

Using the similar-triangle relations directly:

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

Adding,

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

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

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

∴ `z/x + z/y = 1`

Divide by z:

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

Hence proved.