Proof of Law of Sines - Circumcircle of a trangle with its circumcenter on one side of the triangle

						var board = JXG.JSXGraph.initBoard('box1', {
boundingbox: [-10, 6, 10, -14],showNavigation:false, showcopyright: false, axis: false,
});

var p1 = board.create('point', [-6.4, -1.6],{face:'',label:{offset:[-15,0]}});
var p2 = board.create('point', [6.5, -1.6],{face:'',label:{offset:[10,0]}});
var mp1 = board.create('midpoint', [p1, p2],{name:'O',face:''});
var c1 = board.create('circle', [mp1, p2]);
var p3 = board.create('glider', [1.7, 5, c1],{face:'',label:{offset:[0,8]}});
var l1 = board.create('segment', [p1, p2]);
var l2 = board.create('segment', [p1,p3],);
var l3 = board.create('segment', [p2,p3],);
var l4 = board.create('segment', [p1,mp1],{name:"R",dash: 2,withLabel:true,label:{offset:[5,-10]}});

var l5 = board.create('segment', [p2,mp1],{name:"R",dash: 2,withLabel:true,label:{offset:[5,-10]}});
var a1 = board.create('angle', [p1, p3, p2],{name:' ',radius:1});

The image appears in the following posts

  1. 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.

Triangle posted 1 year ago

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