# Geometry Construction Circles II

• Sep 19th 2010, 02:32 PM
MATNTRNG
Geometry Construction Circles II
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.
• Sep 20th 2010, 12:08 AM
earboth
Quote:

Originally Posted by MATNTRNG
[FONT=Times New Roman][SIZE=3]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.
2. The line perpendicular to ME through E yields the 2 chords with equal length. (see attachment)
3. Prove that this construction is valid.
• Sep 20th 2010, 04:12 PM
MATNTRNG
Quote:

Originally Posted by earboth
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?
• Sep 21st 2010, 05:01 PM
Quote:

Originally Posted by MATNTRNG
There is no doubt in my mind that this construction is valid. How would I go about proving it?

Hi 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.
• Sep 23rd 2010, 09:29 AM
Quote:

Originally Posted by MATNTRNG
There is no doubt in my mind that this construction is valid. How would I go about proving it?

The previous post is really not suitable as a proof, 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

$[m_1o_1]$ and $[m_2o_2]$ 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.