1. Construct the large circle, and draw a vertical diameter (this is the significance of the ‘right triangle’ because if the triangle has an angle of 90 degrees at the circumference, then the hypotenuse is a diameter). We do not need the rest of the triangle ABC.
2. Draw a horizontal perpendicular from any point on the circumference to the diameter. Using compasses, construct the small circle with radius BB’.
3. Construct the common chord with endpoints Ch1, Ch2, Ch2 being the further from the diameter. The intersection with BB’ is I.
4. Construct a perpendicular from Ch2 to the diameter at Ch2’. Now, BB’ is a radius of the small circle and BCh2 is a radius of the small circle and Ch2Ch2’B’ is a right angle and BB’Ch2 is a right angle, so B-B’-Ch2’-Ch2 is a square. It follows that BCh2 is vertical and parallel to the diameter of the large circle.
5. Extend BCh2 to the other side of the small circle at Ch2’’. Entend the diameter of the large circle in the same direction so that a further perpendicular can be dropped from Ch2’’ to the extended diameter at D. It will be immediately obvious by symmetry that Ch2Ch1 extended will intercept these two lines at D also (in essence we have constructed a mirror image of ABC, translated from BB1 to Ch2Ch2’and flipped vertically).
6. It follows immediately that Ch2D bisects BB’ at I.
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Posted by broll
on 2010-09-25 01:13:03 |
Solution by demonstration
