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:
[tex]\displaystyle \angle JPO\cong \angle KPO[/tex]
And since OK and OJ are perpendicular bisectors:
[tex]m\angle OKP=90^\circ \text{ and } m\angle OJP=90^\circ[/tex]
Therefore:
[tex]\angle OKP\cong \angle OJP[/tex]
By the Reflexive Property:
[tex]OP\cong OP[/tex]
Therefore:
[tex]\Delta OKP\cong \Delta OJP[/tex]
By AAS Congruence.
Hence:
[tex]OK\cong OJ[/tex]
By CPCTC.
Recall that congruent chords are equidistant from the center.
Thus, by converse, chords that are equidistant from the center are congruent.
Therefore:
[tex]MN\cong RS[/tex]
