Question

In the above figure, $AB\parallel MN\parallel PC$, then which of the following will be true?

• A. $\frac{1}{MN}+\frac{1}{PC}=\frac{1}{AB}$
• B. $\frac{1}{AB}+\frac{1}{MN}=\frac{1}{PC}$
• C. $\frac{1}{AB}+\frac{1}{PC}=\frac{1}{MN}$
• D. $\frac{1}{AB}-\frac{1}{PC}=\frac{1}{MN}$

Answer 1

The correct option is C

$\frac{1}{AB}+\frac{1}{PC}=\frac{1}{MN}$

In $△MNCand△ABC$
$\angle MNC=\angle ABC={90}^{\circ }$
$\angle MCN=\angle ACB\text{(Common angle)}$

Therefore, $△MNC\sim △ABC$ by AA similarity

So, $\frac{MN}{AB}=\frac{NC}{BC}$ ---- I

In $△MNBand△PCB$
$\angle MNB=\angle PCB={90}^{\circ }$
$\angle MBN=\angle PBC\text{(Common angle)}$

Therefore, $△MNB\sim △PCB$ by AA similarity.
So, $\frac{MN}{PC}=\frac{BN}{BC}$ --- II

Adding I and II,
$\frac{MN}{AB}+\frac{MN}{PC}=\frac{NC}{BC}+\frac{BN}{BC}$

$MN[\frac{1}{AB}+\frac{1}{PC}]=\frac{BC}{BC}$
$\Rightarrow \frac{1}{AB}+\frac{1}{PC}=\frac{1}{MN}$