Chapter 15.3, Problem 20PSC

To determine

To Prove: In the given figure segment AB is closer to the center than the segment BC.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Chords }\overline{AB}\text{ and }\overline{BC}\\ AB>BC\end{array}$

  

Formula Used:

  $\begin{array}{l}\text{Pythagorean-Theorem:}\\ {\text{h}}^2=p^2+b^2\end{array}$

Calculation:

  

  $\begin{array}{l}\text{⇒}AB>BC\\ \text{⇒}MB>NB\\ \text{⇒}\text{In }\Delta OMB\text{ and }\Delta ONB\\ \text{⇒}{r}^2=M{B}^2+O{M}^2=O{N}^2+N{B}^2\\ \text{⇒}ON>OM\left(\text{Equation i}\right)\\ \text{⇒}\text{The }\overline{AB}\text{ is closer to the centre }\left(\text{From equation i}\right)\end{array}$

Conclusion:

Thus from the calculation it is proved that $\overline{AB}$ is closer to the centre