prove that the line joining the mid point of two parallel chords of a circle passes through the centre.
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Recall one important thing when you are solving this question:
The perpendicular bisector of any chord passes through the center of the circle.
So when you construct the diagram, remember to mark the right angles (∠OAZ and ∠OBX in the diagram).
Also, you need to identify what you are supposed to prove (joining mid point passes through the center). So you need to prove that AOB is a straight line.
To do that, we can construct another line OC//WX//YZ.
The proof
∠BOC = ∠OAZ = 90° (corresponding angles, OC//YZ)
∠COA = ∠XBO = 90° (corresponding angles, OC//WX)
∠AOB
= ∠BOC + ∠COA
=180°
So AOB is a straight line. Therefore AB passes through O.
The perpendicular bisector of any chord passes through the center of the circle.
So when you construct the diagram, remember to mark the right angles (∠OAZ and ∠OBX in the diagram).
Also, you need to identify what you are supposed to prove (joining mid point passes through the center). So you need to prove that AOB is a straight line.
To do that, we can construct another line OC//WX//YZ.
The proof
∠BOC = ∠OAZ = 90° (corresponding angles, OC//YZ)
∠COA = ∠XBO = 90° (corresponding angles, OC//WX)
∠AOB
= ∠BOC + ∠COA
=180°
So AOB is a straight line. Therefore AB passes through O.
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Step-by-step explanation:
I hope this diagram will help you
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