Centroid (G)
Centroid is the point of intersection of the three medians of a triangle. In the following figure, the three red lines are the medians of triangle ABC:
Median is the line connecting a vertex to the midpoint of the opposite side. In the figure above, Mb is the midpoint of side AC, Ma is the midpoint of side BC, and Mc is the midpoint of side AB.
Since Mc e Mb are midpoints of the sides of triangle ABC, McMb is parallel to side BC, and its measure is 0,50(BC).
R is the midpoint of segment BG, and S is the midpoint of segment CG. Therefore, RS is parallel to side BC, and measures 0,50(BC).
Since McMb and RS are both parallel to side BC, they are also parallel to each other, and measure the same (0,50*BC). Thus McMbSR is a parallelogram, whose diagonals meet at their midpoints (parallelogram property).
So we have:
McG = GS = SC. The Centroid (G) divides the mediam McC in a 2:1 ratio.
MbG = GR = RB. The Centroid (G) divides the median MbB in a 2:1 ratio.
The same line of reasoning applies to median MaA, leading to the same conclusion: all three medians meet at point G (the Centroid), that divides the medians in a 2:1 ratio.
One interesting property of the Centroid is that it is the center of gravity of the triangle: if we suspend the triangle through its centroid, it stays in balance.
Incenter (I)
The Incenter is the point of intersection of the three internal angle bisectors of the triangle.
In the following figure the three red lines are the internal angle bisectors of triangle ABC:
The angle bisector is the line that splits the angle in half. Thus all points on this line are at the same distance from the two sides that form the angle. Since the Incenter is the point where the three bisectors meet, it is at the same distance from all three sides of the triangle, and is the center of a circle inscribed in the triangle.
Circumcenter (O)
The Circumcenter is the point of intersection of the three perpendicular bisectors of the sides of a triangle.
In the following figure the three red lines are the perpendicular bisectors of the sides of triangle ABC:
The perpendicular bisector is a line at right angle passing through the midpoint of each side.
Therefore, in figure above, any point on perpendicular bisector to side AC is at the same distance from vertices A and C. And any point on perpendicular bisector to side AB is at the same distance from vertices A and B. Since point O is on both perpendicular bisectors, it is at the same distance from vertices A, B and C, and is the center of a circle circumscribed to the triangle..
Orthocenter (H)
Orthocenter is the point of intersection of the three altitudes of a triangle. In the following figure the three red lines are the altitudes of triangle ABC:
In order to prove that the three altitudes do intersect one another in one single point, we have drawn the dotted lines in the figure above. A'B' is parallel to AB, B'C' is parallel to BC, and A'C' is parallel to AC. These three dotted lines form triangle A'B'C'.
Consider now the two parallelograms AC'BC and AB'CB. Note that AC'=BC and AB'=BC; therefore, AC'=AB', and A is the midpoint of side B'C'. That means that the altitude to side BC (red line that passes through vertex A) is also the perpendicular bisector to side B'C'.
If we repeat this process to the other two sides of triangle A'B'C', we'll find that the three altitudes of triangle ABC are also the perpendicular bisectors of the sides of triangle A'B'C'. Since the three perpendicular bisectors of a triangle A'B'C' intersect in one single point (Circumcenter of triangle A'B'C'), the three altitudes of triangle ABC will also intersect in one single point.
Ex-center
In the following figure, the three bisectors (red lines) intersect in Ic, one of the Ex-centers of triangle ABC:
Every triangle has three ex-centers.
Solved SAT Practice Tests
Find some solved SAT Practice Tests in the following link: SAT Important Points of Triangles Questions
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