3 prove that the perpendicular bisector of a chord of a circle aluays passes through the centre.
Answers
Answer:
Let F be the center of the circle in question.
Draw any chord AB on the circle.
Bisect AB at D.
Construct CE perpendicular to AB at D, where D and E are where this perpendicular meets the circle.
Then the center F lies on CE.
The proof is as follows.
Join FA,FD,FB.
As F is the center, FA=FB.
Also, as D bisects AB, we have DA=DB.
As FD is common, then from Triangle Side-Side-Side Equality, △ADF=△BDF.
In particular, ∠ADF=∠BDF; both are right angles.
From Book I Definition 10: Right Angle:
When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.
So ∠ADF and ∠BDF are both right angles.
Thus, by definition, F lies on the perpendicular bisector of AB.
Hence the result.
Step-by-step explanation:
Answer:
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