Question

AB, CD, PQ are perpendicular to BD. AB = x, CD = y amd PQ = Z.
Prove that 1/x + 1/y = 1/z

Answer 1

In the attachments I have answered this problem.

I have applied trigonometric ratios to prove the result.

See the attachment for detailed solution.

Answer 2

Given :

AB ,CD , PQ are perpendicular to BD ,

AB = x , CD = y and PQ = z

Proof :

Consider ∆ABD and ∆PQD

<ABD = <PQD = 90°

<ADB = <PDQ ( common angle )

By A.A similarity ,

∆ABD ~ ∆PQD

PQ/AB = DQ/BD

=> z/x = QD/BD ( c.p.s.t ) ---( 1 )

Consider ∆CDB and ∆PQB

<CDB = <PQB = 90°

<CBD = <PBQ ( common angle )

∆CDB ~ ∆PQB ( A.A similarity )

So , z/y = BQ/BD ----( 2 )

From ( 1 ) and ( 2 ) we get

z/x + z/y = QD/BD + BQ/BD

=> z( 1/x + 1/y ) = ( QD + BQ )/BD

=> z( 1/x + 1/y ) = BD/BD

=> z( 1/x + 1/y ) = 1

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

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