Question

AB,CD, PQ are perpendicular on BD . ab is equal to X CD is equals to Y and PQ is equals to Z then prove that 1 by X + 1/Y = 1/Z​

Answer 1

Question:

AB,CD, PQ are perpendicular on BD . ab is equal to X CD is equals to Y and PQ is equals to Z then prove that 1 by X + 1/Y = 1/Z?

Answer:

Given :

AB ,CD , PQ are perpendicular to BD ,

AB = x , CD = y

and, PQ = z

To Prove:

1 by X + 1/Y = 1/Z

Proof:

In ∆ABD and ∆PQD

<ABD = <PQD = 90°

<ADB = <PDQ ( common angle )

<ADB = <PDQ ( common angle )By A.A similarity ,

<ADB = <PDQ ( common angle )By A.A similarity ,∆ABD ~ ∆PQD

<ADB = <PDQ ( common angle )By A.A similarity ,∆ABD ~ ∆PQDPQ/AB = DQ/BD

<ADB = <PDQ ( common angle )By A.A similarity ,∆ABD ~ ∆PQDPQ/AB = DQ/BD=> z/x = QD/BD ( c.p.s.t ) ---( 1 )

In ∆CDB and ∆PQB

∆CDB and ∆PQB <CDB = <PQB = 90°

∆CDB and ∆PQB <CDB = <PQB = 90°<CBD = <PBQ ( common angle )

∆CDB and ∆PQB <CDB = <PQB = 90°<CBD = <PBQ ( common angle )∆CDB ~ ∆PQB ( A.A similarity )

∆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

From ( 1 ) and ( 2 ) we getz/x + z/y = QD/BD + BQ/BD

From ( 1 ) and ( 2 ) we getz/x + z/y = QD/BD + BQ/BD=> z( 1/x + 1/y ) = ( QD + BQ )/BD

From ( 1 ) and ( 2 ) we getz/x + z/y = QD/BD + BQ/BD=> z( 1/x + 1/y ) = ( QD + BQ )/BD=> z( 1/x + 1/y ) = BD/BD

From ( 1 ) and ( 2 ) we getz/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

From ( 1 ) and ( 2 ) we getz/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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Answer 2

Step-by-step explanation:

Refer to the attached image.

Given : AB\parallel PQ \parallel CDAB∥PQ∥CD

To prove: \frac{1}{x}+\frac{1}{y}=\frac{1}{z}

x

1

+

y

1

=

z

1

Proof:

Consider \Delta DPQΔDPQ and \Delta DABΔDAB

\angle D=\angle D∠D=∠D (Common angle)

\angle ABD=\angle PQD∠ABD=∠PQD (Each 90 degree)

(By AA criteria, which states "In two triangles, if two pairs of corresponding angle