![]() ![]() BE=BD, using the Two Tangent theorem.īEOD is thus a kite, and we can use the kite properties to show that ΔBOD is a 30-60-90 triangle. The circle is inscribed in the triangle, so the two radii, OE and OD, are perpendicular to the sides of the triangle (AB and BC), and are equal to each other. So instead, I will use a longer process, that only relies on things we have already proven using triangle congruency. The centroid of an equilateral triangle lies on the medians, which are also perpendicular to the bases, and splits the medians into two segments measuring ⅓ of the length and ⅔ of the length, respectively.īut to prove those properties of the centroid (like the fact that all three medians do in fact meet at one point, or that the centroid is also the center of the circle) is quite complex and beyond the scope of high school geometry. There are several quick ways to find that length which rely on properties of the centroid - the point where all three medians of the triangle meet. We know a circle is fully defined by the length of its radius, r, so the key here will be to find that - the length of segment OD. ![]() ProblemĪ circle is inscribed in an equilateral triangle with side length x. We can use the properties of an equilateral triangle and a 30-60-90 right triangle to find the area of a circle inscribed in an equilateral triangle, using only the triangle's side length.
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