Question 1201144
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Start with square ABCD. 
{{{
drawing(300,300,-4,4,-4,4,

line(0.85,0,1.15,0),line(-1.15,0,-0.85,0),line(0,0.85,0,1.15),line(0,-1.15,0,-0.85),

line(-1,-1,1,-1),line(1,-1,1,1),line(1,1,-1,1),line(-1,1,-1,-1),

circle(-1,-1,0.03),circle(-1,-1,0.06),circle(-1,-1,0.09),circle(1,-1,0.03),circle(1,-1,0.06),circle(1,-1,0.09),circle(1,1,0.03),circle(1,1,0.06),circle(1,1,0.09),circle(-1,1,0.03),circle(-1,1,0.06),circle(-1,1,0.09),

locate(-1,-1.1,"A"),locate(1,-1.1,"B"),locate(1,1.45,"C"),locate(-1,1.45,"D"),

locate(-4,-3,matrix(1,4,Diagram,is,to,scale))
)
}}}


Form equilateral triangles CDP and BCQ, such that the new points are outside the square.
{{{
drawing(300,300,-4,4,-4,4,

line(0.85,0,1.15,0),line(-1.15,0,-0.85,0),line(0,0.85,0,1.15),line(0,-1.15,0,-0.85),

line(0.63,1.94,0.37,1.79),line(-0.63,1.94,-0.37,1.79),line(1.94,-0.63,1.79,-0.37),line(1.94,0.63,1.79,0.37),

line(-1,-1,1,-1),line(1,-1,1,1),line(1,1,-1,1),line(-1,1,-1,-1),

circle(-1,-1,0.03),circle(-1,-1,0.06),circle(-1,-1,0.09),circle(1,-1,0.03),circle(1,-1,0.06),circle(1,-1,0.09),circle(1,1,0.03),circle(1,1,0.06),circle(1,1,0.09),circle(-1,1,0.03),circle(-1,1,0.06),circle(-1,1,0.09),

locate(-1,-1.1,"A"),locate(1,-1.1,"B"),locate(1,1.45,"C"),locate(-1.2,1.45,"D"),

line(1,1,-1,1),line(-1,1,0,2.73),line(0,2.73,1,1),
line(1,-1,1,1),line(1,1,2.73,0),line(2.73,0,1,-1),

circle(1,1,0.03865),circle(1,1,0.0773),circle(1,1,0.11595),circle(-1,1,0.03865),circle(-1,1,0.0773),circle(-1,1,0.11595),circle(0,2.73,0.03865),circle(0,2.73,0.0773),circle(0,2.73,0.11595),circle(1,-1,0.03865),circle(1,-1,0.0773),circle(1,-1,0.11595),circle(1,1,0.03865),circle(1,1,0.0773),circle(1,1,0.11595),circle(2.73,0,0.03865),circle(2.73,0,0.0773),circle(2.73,0,0.11595),

locate(0,3.23,"P"),locate(2.73,-0.25,"Q"),

locate(-4,-3,matrix(1,4,Diagram,is,to,scale))
)
}}}


Then form triangle APQ shown in red.
{{{
drawing(300,300,-4,4,-4,4,

line(0.85,0,1.15,0),line(-1.15,0,-0.85,0),line(0,0.85,0,1.15),line(0,-1.15,0,-0.85),

line(0.63,1.94,0.37,1.79),line(-0.63,1.94,-0.37,1.79),line(1.94,-0.63,1.79,-0.37),line(1.94,0.63,1.79,0.37),

line(-1,-1,1,-1),line(1,-1,1,1),line(1,1,-1,1),line(-1,1,-1,-1),

circle(-1,-1,0.03),circle(-1,-1,0.06),circle(-1,-1,0.09),circle(1,-1,0.03),circle(1,-1,0.06),circle(1,-1,0.09),circle(1,1,0.03),circle(1,1,0.06),circle(1,1,0.09),circle(-1,1,0.03),circle(-1,1,0.06),circle(-1,1,0.09),

locate(-1,-1.1,"A"),locate(1,-1.1,"B"),locate(1,1.45,"C"),locate(-1.2,1.45,"D"),

line(1,1,-1,1),line(-1,1,0,2.73),line(0,2.73,1,1),
line(1,-1,1,1),line(1,1,2.73,0),line(2.73,0,1,-1),

circle(1,1,0.03865),circle(1,1,0.0773),circle(1,1,0.11595),circle(-1,1,0.03865),circle(-1,1,0.0773),circle(-1,1,0.11595),circle(0,2.73,0.03865),circle(0,2.73,0.0773),circle(0,2.73,0.11595),circle(1,-1,0.03865),circle(1,-1,0.0773),circle(1,-1,0.11595),circle(1,1,0.03865),circle(1,1,0.0773),circle(1,1,0.11595),circle(2.73,0,0.03865),circle(2.73,0,0.0773),circle(2.73,0,0.11595),

locate(0,3.23,"P"),locate(2.73,-0.25,"Q"),

red(
line(-1,-1,0,2.73),line(0,2.73,2.73,0),line(2.73,0,-1,-1)
),

locate(-4,-3,matrix(1,4,Diagram,is,to,scale))
)
}}}
The goal is to prove that triangle APQ is equilateral.


For now let's focus on triangle ABQ.
BQ = BC (since BCD is equilateral)
AB = BC (since ABCD is a square)
AB = BQ (by the transitive property)


The AB = BQ statement then tells us triangle ABQ is isosceles.
The vertex angle is 90+60 = 150 degrees.
The base angles are (180-150)/2 = 30/2 = 15 degrees each.


Through similar logic, you'll find that triangles ADP and PCQ are congruent to isosceles triangle ABQ.
All isosceles triangles mentioned have the same base length
AQ = AP = PQ


Therefore, triangle APQ is equilateral.
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