Consider 3 things for the 9 Point circle.Originally Posted bysuedenation

1)3 midpoints on the triangles sides.

2)3 Feet of the triangle altitude.

3)3 Midpoint of of line joining vertex and orthocenter.

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(a)isosceles

Well(1), the midpoints are all distinct in any triangle, thus so far we have 3.

Now, we go the step (2) notice that the alititude of an isoseles triangle is also a midpoint on the base. Thus, on the base the midpoint and foot of altitude coincide. The other 2 feet CAN be distinct. And finally midpoint of vertex and orthocenter is all distinct. Thus, we have AT LEAST 8 distinct points.

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(b)equilateral.

The midpoints are distinct that gives 3. However, all the feet of the altitudes ARE the midpoints thus the feet all coincide with the midpoints. And again, the midpoint of line joining vertex and orthocenter are all distinnt. Thus, for an equilateral they are exactly 6 points.

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(c)right

Again, the midpoints are all distinct. Now, as for the altitude they all meet at on the right angle. Thus, the two non-right angles coincide while the altitute from the right angle may be distinct from the midpoint of the hypotenuse. That gives at most 2. Finally the midpoints of the line joining the ortocenter (which is right angle over here) and vetrices are the midpoints for the two non right angled sides. And as for the right angle no line is formed because the line starts at orthocenter(right angle) and ends at vertex (right angle) no line no midpoint! Thus, there are no new distinct point for condition (3). Thus, we have at most 5.

Note: If it is an isoseles right triangle we than have 4 otherwise 5.