Ipotesi\!:\ \overline{AB} > \overline{AC}\
Tesi\!:\ A\^CB > A\^BC

1\ -\ \overline{AD} \cong \overline{AC}\ -\ 
Per\ costruzione.\ Possibile\ perchè\ per\ Ip.\ 
\overline{AB} > \overline{AC}

2\ -\ A\^DC \cong A\^CD\ -\ \triangle ACD\ 
isoscele\ per\ (1)

3\ -\ A\^DC > A\^BC\ -\ Perchè\ A\^DC\ è\ 
angolo\ esterno\ a\ \triangle BCD

4\ -\ A\^CB > A\^CD\ -\ Per\ costruzione\ D\ è\ 
interno\ a\ A\^CB

5\ -\ A\^CB > A\^DC > A\^BC\ -\ Per\ (4);\ 
(2)\ e\ (3)\
Q.\ E.\ D.

http://www.profcortese.net

A = (0.09544986289664026, 2.4424468975310063)
B = (5.689670606936318, -2.182782111955113)
C = (-2.1946573305288073, -2.182782111955113)
poli1, c, a, b = Polygon(A, B, C)
α = Angle(B, C, A)
β = Angle(A, B, C)
d = Circle(A, C)
D = Intersect(d, c, 1)
e = Segment(A, D)
f = Segment(A, C)
g = Segment(C, D)
γ = Angle(A, D, C)
δ = Angle(D, C, A)