Question

please solve the problem​

Answer 1

Given :

• AB || PQ || CD
• AB = x units
• CD = y units
• PQ = z units

To prove :

• $\sf{\dfrac{1}{x}\:+\:\dfrac{1}{y}\:=\:\dfrac{1}{z}}$

Solution :

In Δ ABD and Δ PQD,

$\sf{\angle\:ABD\:=\:\angle\:PQD\:}$

$\sf{\angle\:ADB\:=\:\angle\:QDB}$

$\sf{\big[\because\:common\:angle\big]}$

•°• ΔBDA ~ Δ QDP.

$\sf{\dfrac{BD}{QD}\:=\:\dfrac{DA}{DP}\:=\:\dfrac{BA}{QP}}$

$\sf{\big[Corresponding\:sides\:of\:similar\:triangles\:are\:in\:proportion\big]}$

$\sf{\dfrac{BD}{QD}\:=\:\dfrac{x}{z}}$

By invertendo,

$\sf{\dfrac{QD}{BD}\:=\:\dfrac{z}{x}\:\:\:(1)}$

Now, in ΔCDB and ΔPQB,

$\sf{\angle\:CDB\:=\:\angle\:PQB}$

$\sf{\angle\:CBD\:=\:\angle\:PBQ}$

$\sf{\big[\because\:common\:angle\big]}$

•°• Δ DBC ~ Δ QBP.

$\sf{\dfrac{DB}{QB}\:=\:\dfrac{BC}{BP}\:=\:\dfrac{DC}{QP}}$

$\sf{\big[Corresponding\:sides\:of\:similar\:triangles\:are\:in\:proportion\big]}$

$\sf{\dfrac{DB}{QB}\:=\:\dfrac{y}{z}}$

By invertendo,

$\sf{\dfrac{QB}{DB}\:=\:\dfrac{z}{y}\:\:\:(2)}$

Add equation (1) and (2),

$\sf{\dfrac{z}{x}\:+\:\dfrac{z}{y}\:=\:\dfrac{QD}{BD}\:+\:\dfrac{QB}{DB}}$

$\sf{\dfrac{z}{x}\:+\:\dfrac{z}{y}\:=\:\dfrac{QD+QB}{BD}}$

$\sf{\dfrac{z}{x}\:+\:\dfrac{z}{y}\:=\:\dfrac{BD}{BD}}$

$\sf{\dfrac{z}{x}\:+\:\dfrac{z}{y}\:=\:1}$

$\sf{\:z\:=\:\Big(\dfrac{1}{x}\:+\:\dfrac{1}{y}\Big)\:=\:1}$

$\sf{\dfrac{1}{x}\:+\:\dfrac{1}{y}\:=\:\dfrac{1}{z}}$

Answer 2

$\bf{\underline{\underline{Given}}}$

$\mathsf{\star\: \: AB \: II \: PQ \: II \: CD}$

$\mathsf{\star\: \: AB = x \: units}$

$\mathsf{\star\:\: CD = y \: units}$

$\mathsf{\star\:\: PQ = z \: units}$

$\bf{\underline{\underline{To\:\:prove}}}$

$\mathsf{\star\:\: \frac{1}{x} + \frac{1}{y} = \frac{1}{z} }$

$\bf{\underline{\underline{Solution}}}$

$\large{\mathsf{\blue{Let\:\:BQ = a\: \: and\: \: DQ = b}}}$

In ∆ ABD and ∆ PQD

$\mathsf{\angle ADB = \angle PDB \: \: \: (common) }$

$\mathsf{\angle ABQ = \angle PQD \: \: \: (corresponding\: angles) }$

∴ ∆ ABD ~ ∆ PQD (by AA criteria)

$\mathsf{\implies\: \frac{AB}{PQ} = \frac{BD}{DQ}}$

$\mathsf{\implies\: \frac{AB}{BD} = \frac{PQ}{DQ}}$

$\mathsf{\implies\: \frac{AB}{BQ + DQ} = \frac{PQ}{DQ}}$

Let's substitute the value

$\mathsf{\implies\: \frac{x}{a + b} = \frac{z}{b}}$

Taking the reciprocal

$\mathsf{\implies\: \frac{a + b}{x} = \frac{b}{z}}$

On cross multiplying, we get:

$\mathsf{\implies\: z (a + b) = bx}$

$\mathsf{\implies\: az + bz = bx}$

$\mathsf{\implies\: az = bx - bz}$

$\mathsf{\implies\: a = \frac{b (x - z)}{z}\: \: \: \red{\rightarrow \: (1)}}$

In ∆ DBC and ∆ QBP

$\mathsf{\angle DBC = \angle QBP \: \: \: (common) }$

$\mathsf{\angle BDC = \angle BQP \: \: \: (corresponding\: angles) }$

∴ ∆ DBC ~ ∆ QBP (by AA criteria)

$\mathsf{\implies\: \frac{BD}{BQ} = \frac{CD}{PQ}}$

$\mathsf{\implies\: \frac{PQ}{BQ} = \frac{CD}{BD}}$

$\mathsf{\implies\: \frac{PQ}{BQ} = \frac{CD}{BQ + DQ}}$

Let's substitute the value

$\mathsf{\implies\: \frac{z}{a} = \frac{y}{a + b}}$

Taking the reciprocal

$\mathsf{\implies\: \frac{a}{z} = \frac{a + b}{y}}$

On cross multiplying, we get:

$\mathsf{\implies\: ay = z (a + b) }$

$\mathsf{\implies\: ay = az + bz}$

$\mathsf{\implies\: ay - az = bz}$

$\mathsf{\implies\: a (y - z) = bz}$

$\mathsf{\implies\: \frac{b (x - z)}{z} (y - z) = bz \: \: \: \red{[using \:(1)]}}$

$\mathsf{\implies\: \frac{\cancel{b} (x - z)}{z} (y - z) = \cancel{b} z}$

$\mathsf{\implies\: (x - z) (y - z) = z^{2}}$

$\mathsf{\implies\: xy - xz - yz + z^{2} = z^{2}}$

$\mathsf{\implies\: xy - xz - yz + \cancel{z^{2}} = \cancel{z^{2}}}$

$\mathsf{\implies\: xy = xz + yz}$

Dividing them by xyz, we get:

$\mathsf{\implies\: \frac{xy}{xyz} = \frac{xz}{xyz} + \frac{yz}{xyz}}$

$\fbox{\mathsf{\green{\implies\: \frac{1}{z} = \frac{1}{y} + \frac{1}{x}}}}$

Hence Proved