Final Answer:
The line joining the midpoints of two parallel chords in a circle passes through the center of the circle.
Step-by-step explanation:
Geometric Proof:
Let O be the center of the circle, let AB and CD be two parallel chords, and let M and N be the midpoints of AB and CD, respectively. We want to show that the line MN passes through O.
Draw the radii OA, OB, OC, and OD. Since M and N are the midpoints of AB and CD, respectively, we have AM = MB and CN = ND.
Since AB is parallel to CD, we have ∠AOB = ∠COD.
Since OA and OB are radii of the circle, we have ∠AOB = 90°.
Therefore, ∠COD = 90°.
Triangles AOM and CON are congruent by the Right-Angle-Side-Side (SAS) Congruency Theorem:
∠AOM = ∠CON (both are right angles)
AO = CO (both are radii of the circle)
AM = CN (given)
Therefore, we have OM = ON.
Since OM = ON and M and N are distinct points, the line MN passes through the center O of the circle.
Algebraic Proof:
Let A, B, C, and D be the points on the circle, and let M and N be the midpoints of AB and CD, respectively. Let O be the center of the circle, and let x be the distance from O to A.
Since M is the midpoint of AB, we have AM = MB = x.
Since N is the midpoint of CD, we have CN = ND = y.
Since MN is parallel to AB and CD, we have MN || DC.
Therefore, the angles between corresponding sides are equal, so we have ∠MON = ∠MOC.
Since O is the center of the circle, we have ∠MOC = 90°.
Therefore, ∠MON = 90°.
Since triangles MON and MOC are right triangles with a common angle, we have MO || NO.
Since MO || NO and M and N are distinct points, the line MN passes through the center O of the circle.