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Given two circles, 1 and 2, with centers at A and B. Let E be one of the two points where the circles intersect. Construct a line FG, through E, on which the two circles cut out equal chords, where neither F nor G are at the other intersection point of the two circles.
1. Construct the midpoint M of AB.Quote:
Originally Posted by MATNTRNG
2. The line perpendicular to ME through E yields the 2 chords with equal length. (see attachment)
3. Prove that this construction is valid.
There is no doubt in my mind that this construction is valid. How would I go about proving it?Quote:
Originally Posted by earboth
Hi MATNTRNG,Quote:
Originally Posted by MATNTRNG
One way is to show that if we draw 2 more circles,
same radii as Circle1 and Circle 2, but centred on the original circle's centres as shown,
the geometry is entirely symmetrical since triangle FmG is isosceles,
due to the two chords being of equal length.
We can repeat the geometry under the centreline,
hence point "m" is on the centreline,
and due to the left-right symmetry, also must be midway between the circle centres.
The previous post is really not suitable as a proof, MATNTRNG.Quote:
Originally Posted by MATNTRNG
Instead, the attachment shows that if the chords are to be of equal length,
then they form the basis of numerous isosceles triangles, two of which are shown.
Since the chords themselves form isosceles triangles with the circle centres, then
and
are parallel.
These line segments are also parallel to the perpendicular line passing through
the point at which the chords meet.
The point on the joint-centreline of the two circles that is equidistant from F and G lies on that perpendicular line
and therefore is midway between the centres.
