### Radical Axis and Center, an Application

The applet below presents a problem from *Mathematical Miniatures*, an exciting book by MAA.

Let points C and D lie on a semicircle with diameter AB. Points E, F, G are the midpoints of chords AC, CD, and DB, respectively. Let the perpendicular from E to AF intersect the (vertical) tangent to the semicircle at A in M, and the perpendicular from G to BF intersect the tangent at B in N. Then MN is parallel to CD.

What if applet does not run? |

### References

- S. Savchev, T. Andreescu,
*Mathematical Miniatures*, MAA, 2003, pp. 114-115

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### Radical Axis and Center, an Application

What if applet does not run? |

The quadrilateral OEFG is the Varignon parallelogram of the quadrilateral ACDB. Its diagonals OF and EG bisect each other in, say, O_{3}. Denote the original (semi)circle as c and that with diameter EG as c_{3}. I want to show that MN serves as the radical axis of the two circles. Since, the center line OO_{3} is perpendicular to the radical axis of the two circles, and coincides with OF, which is perpendicular to CD, this would imply that MN is indeed parallel to CD.

Introduce circles c_{1} and c_{2} with diameters AO and OB, respectively. E is the midpoint of the chord AC in c, so that EO is perpendicular to AC. Thus, the angle AEO is right, and E lies on c_{1}. Similarly, G lies on c_{2}.

O_{1}O_{3} is the midline in ΔAOF and is therefore parallel to AF. By the construction, the center line O_{1}O_{3} is then perpendicular to EM. But so is the radical axis of c_{1} and c_{3}. Since both pass through the points of intersection (E, in particular) of c_{1} and c_{3}, we may claim that EM is the radical axis of c_{1} and c_{3}. On the other hand, the radical axis of c and c_{1} is their common tangent at A. The two axes intersect at the radical center of c, c_{1} and c_{3}, which is M. Being the radical center of the circle triad c, c_{1} and c_{3}, M lies on the radical axis of c and c_{3}. In a similar fashion, N is the radical center of the triad c, c_{2} and c_{3} and, as M, also lies on the radical axis of c and c_{3}. We see that MN is in fact the radical axis of c and c_{3}, which concludes the proof.

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