Question

In the following figure, AB, CD and EF are perpendicular to the straight line BDF.

If AB = x units and CD = z units and EF = y units, then which of the following is/are true?

• A. $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$
• B. $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$
• C. $\frac{1}{x}-\frac{1}{z}=\frac{1}{y}$
• D. $-\frac{1}{x}-\frac{1}{z}=\frac{1}{y}$

Answer 1

The correct option is A

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

In the given figure,

AB, CD and EF are perpendicular to the line BDF.

AB = x, CD = z, EF = y

Let BD = a and DF = b

In $\Delta ABF,and\Delta CDF,$

$CD||AB$

$\angle CFD=\angle AFB$ (Common angle)

$\frac{CF}{AF}=\frac{DF}{BF}$ (by BPT)

$\Delta ABF\sim \Delta CDF$ (by SAS simlarity criterion)

Hence, $\frac{CF}{AF}=\frac{DF}{BF}=\frac{CD}{AB}$.

$\Rightarrow \frac{z}{x}=\frac{b}{a+b}$ ...(i)

Similarly in $\Delta BEFand\Delta BCD,$

$CD||EF$

$\angle CBD=\angle EBF$ (Common angle)

$\frac{BC}{BE}=\frac{BD}{BF}$ (by BPT)

$\Delta BEF\sim \Delta BCD$ (by SAS simlarity criterion)

$\frac{BC}{BE}=\frac{BD}{BF}=\frac{CD}{EF}$

$\Rightarrow \frac{z}{y}=\frac{a}{a+b}...\left(ii\right)$

Adding (i) and (ii)

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

$\Rightarrow z(\frac{1}{x}+\frac{1}{y})=1\Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{1}{z}$