Answer:
See Below.
Step-by-step explanation:
Please refer to the diagram below.
We are given that O is the center of the circle, and chords MN and RS are intersected a P. OP is the bisector of ∠MPR. And we want to prove that MN = RS.
We will construct segments OK and OJ such that it perpendicularly bisects MN and RS.
Since OP bisects ∠MPR, it follows that:
![\displaystyle \angle JPO\cong \angle KPO](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cangle%20JPO%5Ccong%20%5Cangle%20KPO)
And since OK and OJ are perpendicular bisectors:
![m\angle OKP=90^\circ \text{ and } m\angle OJP=90^\circ](https://tex.z-dn.net/?f=m%5Cangle%20OKP%3D90%5E%5Ccirc%20%5Ctext%7B%20and%20%7D%20m%5Cangle%20OJP%3D90%5E%5Ccirc)
Therefore:
![\angle OKP\cong \angle OJP](https://tex.z-dn.net/?f=%5Cangle%20OKP%5Ccong%20%5Cangle%20OJP)
By the Reflexive Property:
![OP\cong OP](https://tex.z-dn.net/?f=OP%5Ccong%20OP)
Therefore:
![\Delta OKP\cong \Delta OJP](https://tex.z-dn.net/?f=%5CDelta%20OKP%5Ccong%20%5CDelta%20OJP)
By AAS Congruence.
Hence:
![OK\cong OJ](https://tex.z-dn.net/?f=OK%5Ccong%20OJ)
By CPCTC.
Recall that congruent chords are equidistant from the center.
Thus, by converse, chords that are equidistant from the center are congruent.
Therefore:
![MN\cong RS](https://tex.z-dn.net/?f=MN%5Ccong%20RS)