# Solving for the radius given a circumscribed triangle

Given an equilateral triangle with it's center indicated by the $\bullet$, I can drop a perpendicular bisector $y$. How can I show the segment connecting the top of y to the bottom right corner bisects the $60^{\circ}$ angle?

The problem I am actually trying to solve has a circumscribed equilateral triangle and I am trying to solve for the radius (the aforementioned segment).

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Assuming the known variable to be the side of the triangle, I am proceeding with my answer.

Let the side of the triangle be 'a'.

1) In an equilateral triangle the median, perpendicular bisector, altitude, angular bisector are the same straight line. Hence the line joining the right most corner and passing through the dot(top of Y) is the angular bisector for that angle. Hence the angle is $30$ degrees.

2)So now you can easily solve for the cir-cum radius, which is the line joining the right most end to the dot(top of Y). You can also find the In-radius which the segment 'y'.

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I've come up with a way to show this.

Let a side of the triangle be $d$.

The segment dropped down from the center, $y$, bisects the base of the triangle into $d/2$, $d/2$.

Let $r$ be a segment which extends diagonally from the center to intersect with the angle in the right corner into, $\theta$, $60^{\circ} - \theta$.

From the right triangle, we know $90^{\circ} - \theta + 60^{\circ} - \theta = 180^{\circ} - 90^{\circ} \implies \theta = 30^{\circ}.$

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Actually I am not sure how to justify my assertion that 90∘−θ+60∘−θ=180∘−90∘. – dal102 Mar 27 '13 at 6:57

Center of the triangle means its Circumcentre

Note: In any equilateral triangle. Centroid,orthocentre, circumcentre and incentre coincide. The circumcentre of an equilateral triangle is also an incentre of the triangle and since the meeting of angular bisectors is incentre, a line drawn from the centre of triangle to any of the vertices bisect the angle.

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