Nine point circle of a triangle ABC is a circle passing through the mid points of sides of the triangle (G, H, I), feet of the perpendicular (D, E , F) drawn from the vertex to the opposite sides and the mid-points (K , L, M) of the distance from the orthocenter (N) to the three vertices of the triangle.
The concept of a nine point circle can be generalized to a nine point ellipse or a nine point hyperbola if we consider a general cevian instead of altitude. A cevian is any segment drawn from the vertex of a triangle to the opposite side. Cevians with special properties include altitudes, angle bisectors, and medians.
Consider three concurrent cevians with cevian point P, locate mid-points E, F and G of the segments from cevian point to the vertices of the triangle. Also locate the feet of the cevian K , N and L. If we draw a conic through any five of the above points, we will get an ellipse and it will also pass through the sixth point.
Now locate the mid points of the three sides of the triangle, we will find that these points also fall on the ellipse constructed above. The conic remains an ellipse when the feet of cevians lie on the sides of the triangle and converts to a nine point hyperbola when the feet of the cevians lie on the extensions of the sides.
Now locate centroid (G1) of the triangle ABC and centre (N1) of the conic , interestingly , the cevian point P , G1 and N1 lie on the same straight line with N1P = 3 N1G1 . This is also the generalization of Euler Line.
Consider three concurrent cevians with cevian point P, locate mid-points E, F and G of the segments from cevian point to the vertices of the triangle. Also locate the feet of the cevian K , N and L. If we draw a conic through any five of the above points, we will get an ellipse and it will also pass through the sixth point.
Now locate the mid points of the three sides of the triangle, we will find that these points also fall on the ellipse constructed above. The conic remains an ellipse when the feet of cevians lie on the sides of the triangle and converts to a nine point hyperbola when the feet of the cevians lie on the extensions of the sides.
Now locate centroid (G1) of the triangle ABC and centre (N1) of the conic , interestingly , the cevian point P , G1 and N1 lie on the same straight line with N1P = 3 N1G1 . This is also the generalization of Euler Line.
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ReplyDeleteIn geometry, the nine-point conic of a complete square is a conic that moves into the three diagonal cases and the six midpoints of surfaces of the complete forum. Use of Content writing service to gets the best writing service at low rates. The better-known nine-point circle is an example of Boucher's conic.
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