You can’t use that theorem about the perpendicular bisector, because it doesn’t say that there is *no other line* through the center; you would have to show that it is a bisector too.

I am sure there are many other ways to come to this conclusion; one thing I like about these problems is that many of them require multiple steps, and that leaves a lot of room for creativity. But no formal proof is required.

]]>For Problem 64, is x 90 degrees?

(In case you’d like to see HOW I got 90 degrees, I did

BĚD = 60° (Corresponding Angles)

BĚD = OĎE (OE = OD = radii)

EÔD = 60° (Angles in a triangle)

CÔD = 60° (Angles on a straight line)

If BC is drawn

OBC = (180-60)/2 = 60° (OB = OC = radii)

ABO =90 – 60 = 30° (Angles in a semicricle)

AÔF = 60° (Vertically Opposite)

FÔB = 60° (Angles on a straight line)

Therefore x = 180 – (30+60) = 90

Alternatively, there’s the theorem which states that the perpendicular bisector of a chord passes through the centre of the circle, so x is automatically 90 degrees.)

Is this correct?

Thanks

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