Question

How should I solve this?

Answer 1

The parallel sides AB, PQ, and CD, gives similar triangles, ∆ABD ~ ∆PQD and ∆CDB ~ ∆PQB, from which we have;

$\frac{1}{x} + \frac{1}{y}= \frac{1}{z}$

Which method can be used to prove the given relation?

From the given information, we have;

• ∆ABD ~ ∆PQD
• ∆CDB ~ ∆PQB

According to the ratio of corresponding sides of similar triangles, we have;

$\frac{x}{z} = \mathbf{\frac{BD}{QD} }$

$\frac{y}{z} = \frac{BD}{ BQ}$

Which gives;

$\mathbf{\frac{y}{z}} = \frac{BD }{ BD - Q D}$

$\frac{z}{y} = \frac{BD - QD }{ BD } = 1 - \frac{Q D }{ BD}$

QD × x = BD × z

BD × z = (1 - QD/BD) × y = y - (QD × y/BD)

Therefore;

BD × z = y - (QD × y/BD)

BQ × y = y - (QD × y/BD)

BQ × y = y - (z × y/x) = y × (1 - z/x)

(1 - z/x) = BQ

BD × z = y × (1 - z/x)

BD = (y × (1 - z/x))/z

Therefore;

QD × x = y × (1 - z/x)

(BD-BQ) × x = y × (1 - z/x)

(BD-(1 - z/x)) × x = y × (1 - z/x)

BD = (y × (1 - z/x))/x + (1 - z/x)

BQ + QD = (1 - z/x) + (y × (1 - z/x))/x

BD = BQ + QD

(y × (1 - z/x))/x + (1 - z/x) = (y × (1 - z/x))/z

(1 - z/x)×(y/x + 1) =(1 - z/x) × y/z

Dividing both sides by (1 - z/x) gives;

y/x + 1 = y/z

Dividing all through by y gives;

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

• 1/x + 1/y = 1/z

Therefore;

$\frac{1}{x} + \frac{1}{y}= \frac{1}{z}$

Learn more about characteristics similar triangles here:

https://brainly.com/question/1799826

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