![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjq8NgxekoyL_LBond010vJ_RSQCc4u43PW6a5HNc7SoQciE64XDRTu0zD2zGCDa8adRY5k1oxSnzvcR6Vwh66NVzfUPuzwiL-VQAuHZdwVYsiqlg1UmqaKXz7uYsdrjQ0oGlq1FUkb4Ns/s400/nedialtri2.jpg)
I wrote recently about the concept of a Nedian; an extension of the idea of medians created by Professor John Satterly of the University of Toronto. I have played around a little more with the concept and come up with a calculation for the area of what I call a nedial triangle, extending the idea of a medial triangle.
Most High School students are introduced to the Medians of a triangle, and it is quite easy to show using basic high school geometry that the triangle whose vertices are the three feet of the medians has an area equal to 1/4 the area of the original triangle. In fact, it is pretty easy to show that the original triangle can be dissected into four congruent copies of the medial triangle.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhDfb783LVEm4sqJWxHJSRyJwm-qoRXfc60rgIt3Pw27CCBFg6Uj6koefliDxPGXFwfYeLwXmyCk1uAx-fM56ttSQ8_3Jsz-qqBKt0bsOIAaXiKTpUwLbShD58r1lKzpHWd_-5xpweGvXU/s400/nedialtri.jpg)
If instead we use the feet of the three nedians, we get a triangle I have called the nedial triangle. I have worked out that the area of such a triangle for an n-nedian (the cevian that cuts the opposite side 1/n th of the way along the edge) will have an area of
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhgCUPTK3WOC2Z3SBa0cPB7xBiqiKEoie1p3M-AJGMiiosGF7tyu7pFUQuqrSmsVq-lmIRuTY79GoFaSIAGA-np6SqamgQYbQWNM51V58LuUvQbCQmjiSAtA9nbOJF98YNKCL3pT1SPjwg/s400/nedialtri3.jpg)
The image below shows the 4-nedial triangle, which has an area of 7/16 the area of ABC.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjq8NgxekoyL_LBond010vJ_RSQCc4u43PW6a5HNc7SoQciE64XDRTu0zD2zGCDa8adRY5k1oxSnzvcR6Vwh66NVzfUPuzwiL-VQAuHZdwVYsiqlg1UmqaKXz7uYsdrjQ0oGlq1FUkb4Ns/s400/nedialtri2.jpg)
The area of the three triangles at the vertices of the original triangle will each be
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjPF2fQwBFjfaQRCy-M1ZF8-baa9D1OnbqmXtLqp-FuS2Bd0y1FFr2lCVY2Vb6QITZ3R4ZYP58AZWCozESSKJl84FBN9fpE2ixf5lJNKIxfkj26vCvz6dPRlZHqLjHMmwtRPocK7F5dP_k/s400/nedialtri4.jpg)
The area of the nedial triangle will grow from a minimum of 1/4 the area of ABC increasing toward a limit of one as n approaches infinity. This can be confirmed by L'Hopital's rule or simple division of the terms. The fact that the limit is also one as n approaches one points out that when the base of the nedian is more than half way along the opposite side and the "n" in the nedian must be less than 2 but more than one we are measuring the area of what Satterly called the "backward nedians" which will create a nedial triangle of the same area. n=3/2 gives the exact area of n=3 (although the triangles are NOT congruent). In general for every k-nedial the area given by the nedial with index equal to n/n-1 has the same area.
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