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I have a triangle here, how do I prove that $BCD$ is equilateral(so all lines have the same length)

And yes this is 2D

Triangle

What I have so far is $$BAC = 120^\circ$$

So how do I point out that $$BCD = 60^\circ$$ $$CBD = 60^\circ$$ $$BDC = 60^\circ$$ Where is the relationship ?

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    $\begingroup$ $\angle CDB=60^o$ due to it being opposite to $\angle CAB=120^o$ and $\angle BCD=\angle BAD=60^o$ since the line $BD$ is common to triangles $BCD$ and $BAD$. $\endgroup$ – K. Rmth Dec 12 '13 at 22:30
  • $\begingroup$ @BrianM.Scott you're right, edited $\endgroup$ – Mazzy Dec 12 '13 at 22:33
  • $\begingroup$ @Mazzy : you can get a degree symbol using ^\circ $\endgroup$ – Stefan Smith Dec 13 '13 at 0:53
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c = 120 due inscribed angle BAD. in the same way arc CD = c

so, CBD = BCD which is 60.

in the same way as arc CB = 240 and arc CB = 120. CDB = 60

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Angle BAD and angle BCD share the chord in the right way, so they are equal. The same for angle CAD and angle CBD. This gives us two angles of the triangle BCD equal to 60 degrees, so that's it.

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  • $\begingroup$ What do you mean by share the chord in the right way? $\endgroup$ – Mazzy Dec 12 '13 at 22:37
  • $\begingroup$ Lay on the same side of it. E.g. CAB and CDB do not share the chord CB 'in the right way', because they are at different sides of it. $\endgroup$ – Polydarya Dec 12 '13 at 22:40
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Firtly, since A、B、C、D are concyclic, we can get ∠CAB+∠CDB=180°, therefore ∠CDB=60°.

Secondly, since ∠CAD=∠BAD, we can get arc CD=arc BD, therefore CD=BD, and ∠DCB=∠DBC.

Since ∠DCB+∠CBD+∠BDC=180°, we can get ∠DCB=∠CBD=∠BDC=60°.

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You first note that arcs, or line segments of equal lengths subtend equal angles. Then $\angle CBD = \angle CAD$, as they are subtended by the same line segment $CD$. Then in same fashion $\angle BCD = \angle BAD$ as they are subtended by the same line segment $BD$. Then you are done as the sum of angles is $180^\circ$ within a triangle.

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