Answer:
See Below.
Step-by-step explanation:
We are given two intersecting circles with centers M and N.
And we want to prove that: I) PQ ⊥ MN and that II) PR = RQ.
Since MP and MQ are radii of the same circle:
[tex]MP\cong MQ[/tex]
Likewise, since NP and NQ are radii of the same circle:
[tex]NP\cong NQ[/tex]
And by the Reflexive Property:
[tex]PQ\cong PQ[/tex]
Therefore, by SSS Congruence:
[tex]\Delta MPN\cong \Delta MQN[/tex]
By CPCTC:
[tex]\displaystyle \angle PMN\cong \angle QMN[/tex]
And by the Reflexive Property:
[tex]MR\cong MR[/tex]
And since they are the radii of the same circle:
[tex]MP\cong MQ[/tex]
Therefore, by SAS Congruence:
[tex]\Delta MPR\cong \Delta MQR[/tex]
Therefore, by CPCTC:
[tex]PR\cong RQ[/tex]
Note that PQ is a chord in Circle M.
Therefore:
[tex]\text{Segment $MN$ bisects chord $PQ$}[/tex]
In a circle, a segment that passes through the center of the circle that is perpendicular to a chord also bisects the chord.
And conversely, a segment that passes through the center of a circle that bisects a chord in the circle is also perpendicular to the chord.
So:
[tex]\displaystyle PQ\perp MN[/tex]
