﻿ Proof the ratio of a side and the sines of its opposite angle of a triangle is equal to the diameter of the circumscribed circle of the triangle

# Proof the ratio of a side and the sines of its opposite angle of a triangle is equal to the diameter of the circumscribed circle of the triangle

The above statement is actually the law of sines, which tells us that the lengths of the sides of a triangle over the sines of their corresponding opposite angles are equal to a constant variable, i.,e., the diameter of the circumscribed circle of the triangle.

Considering three cases when the circumcenter is inside or outside the triangle, or on one side of the triangle, we will prove the equation, for any given side c and its opposite interior angle C in a triangle,

\dfrac{c}{\sin C} =2R ,

in which R is the radius of the circumcircle of the triangle.

## The circumcenter is inside the triangle

In this case, angle C and the central angle ∠AOB intersect the arc, so

∠C = \dfrac{1}{2} ∠AOB

As △AOB is an isosceles triangle, ∠AOD is the bisector of ∠AOB ,

therefore,

∠AOD = ∠C , \sin C = \sin∠AOD

\sin∠AOD= \dfrac{\dfrac{c}{2} }{AO} = \dfrac{c}{2R}

\therefore \dfrac{c}{\sin C} =2R

## The circumcenter is outside the triangle

In this case, ∠C is an obtuse angle, so the central angle intersecting the arc ∠AOB is an angle more than 180 \degree.

According to the Inscribed Angle Theorem,

∠C = \dfrac{1}{2} ∠AOB =∠AOG

\because ∠AOF = 180 \degree - ∠AOG

\therefore \sin ∠AOF = \sin ∠AOG= sin ∠C

\therefore sin ∠C = = \dfrac{\dfrac{c}{2} }{AO} = \dfrac{c}{2R}

\therefore \dfrac{c}{\sin C} =2R

## The circumcenter is on one side of the triangle

In this case,

∠C = 90\degree

and

c=2R

\dfrac{c}{\sin C} =\dfrac{2R}{\sin 90\degree } = 2R

## Summary

In summary, there are 3 cases based on the position of the circumcenter and the inscribed triangle,

1. the circumcenter is inside the triangle
2. the circumcenter is outside the triangle
3. the circumcenter is on one side of the triangle

The types of the triangle is also determined in he three cases , which are acute triangle, obtuse triangle and right triangle. In any of the cases, it's true that the ratio of any side and the sines of its opposite angle is equal to the diameter of the circumscribed circle.

Steven Zheng posted 3 years ago

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