Equilateral triangle and 120° angle
P is the angle's vertex. With A,B we form an equilateral triangle on the same direction the angle's rays open up.
How can I prove the triangle's third vertex C is always on the bisector's line?
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$\begingroup$As pointed out by some comments, the problem has been originally misstated. It should be: we are given a 120 degrees angle with vertex P and an equilateral triangle ABC inscribed in it, such that the vertices A and B lie on the sides of the angle and C is in its interior, as in the drawing. Show that the line PC bisects the angle APB (i.e., that the angles APC and CPB are both 60 degrees).
Proof. Consider the quadrangle ACBP. It is cocyclic (its vertices lie on the same circle) since the sum of the angles at its opposite vertices P and C is 180 degrees. Thus the angles APC and CPB are equal, since they are inscribed angles in the same circle that are facing equal segments. QED
From the proof you can see that it is enough to assume that ABC is isosceles, i.e. AC=BC, and that the sum of the angles and P and C is 180 degrees.
$\endgroup$ 11 $\begingroup$Draw two perpendicular lines from C to PA and PB, and the intersecting points are D and E, as shown below:
What we need to prove is that △CDA≌△CEB, therefore CD=CE, and since any point on an angle bisector has equal distance to the two side of the angle, we finish proving that C is on the angle bisector of ∠APB.
It is easy to get that CA=CB, ∠CDA=∠CEB=90∘. We need to get one more condition, which is ∠CAD=∠CBE.
Apparently, ∠CAB=∠CPB=60∘, therefore C、A、P、B are concyclic. Thus, ∠ACP=∠ABP. We can see that ∠CAD=∠CPA+∠ACP=60∘+∠ABP=∠CBA+∠ABP=∠CBE.
$\endgroup$ $\begingroup$Call $\;M\;$ to the intersection point of $\;PC\,,\,AB\;$ , and observe that
$$\Delta APM\sim\Delta BPM\;\;\text{by s.a.s., since applying the angle bisector theorem we get}$$
$$\frac{AM}{MB}=\frac{AP}{PB}\;,\;\;\text{on the triangle}\;\;\Delta APB$$
This means $\;PC\perp AB\;$ (why?) , so both $\;\Delta CBP\;,\;\Delta CAP\;$ are right triangles (and, in fact, golden ones $\;30-60-90\;$).
Now just remember that an angle's bisector is the locus of all points on the plane that are equidistintant from both angle's rays.
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