Ipotesi\!:\ \overline{AC} \cong \overline{BC};\ 
\overline{AD} \cong \overline{BE};\ 
\overline{HD} \cong \overline{EK} 

Tesi\!:\ \overline{AH} \cong \overline{BK};\ 
\triangle ABT\ isoscele 

Ragionamento\!:\ la\ tesi\ è\ dimostrata\ se 
\triangle BEK \cong \triangle ADH;\
se\ dimostriamo\ che\ B\^EK \cong 
A\hat{D}H\ allora...\
quindi\!:

1\ -\ C\^AB \cong C\^BA\ - Angoli\ alla\ base\ 
di\ triangolo\ isoscele

2\ -\ \triangle ABE \cong \triangle BAD\ -\ 
I\ criterio\!:\ \overline{AB}\ comune;\ 
\overline{AD} \cong \overline{BE}\ (Ip.);\
\ \ \ \ \ \ \ D\^AB \cong E\^BA\ (1)

3\ -\ B\^EA \cong A\^DB\ -\ Per\ (2)

4\ -\ B\^EH \cong A\^DH\ -\ Supplementari\ di\ angoli\ congruenti\ (3)

5\ -\ \triangle BEK \cong \triangle ADH\ -\ 
I\ criterio\!:\ \overline{BE} \cong \overline{AD}\ 
(Ip.);\
\ \ \ \ \ \ \ \overline{EK} \cong \overline{DH}\ 
(Ip.);\ B\^EK \ \cong A\^DH\ (4)

6\ -\ \overline{BK} \cong \overline{AH}\ -\ Per\ (5)\ 
-\ Prima\ parte\ della\ tesi

7\ -\ E\^BK \cong D\^AH\ -\ Per\ (5)

8\ -\ T\^BA \cong T\^AB\ -\ Supplementari\ di\ angoli\ congruenti;\ (1);\ (7)

9\ -\ \triangle ABT\ isoscele\ -\ Angoli\ alla\ base\ congruenti;\ (8)\
Q.\ E.\ D.

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A = (-1.02, -1.62)
B = (3.06, -1.62)
c = Segment(A, B)
r = LineBisector(c)
C = Point(r)
a = Segment(B, C)
b = Segment(A, C)
D = Point(b)
E = Intersect(Circle[B, Segment[A, D]], a)
d = Segment(A, D)
e = Segment(B, E)
f = Ray(A, E)
K = Point(f)
g = Segment(A, E)
h = Segment(E, K)
i = Ray(B, D)
j = Segment(B, D)
H = Intersect(Circle[D, h], i, 2)
k = Segment(D, H)
l = Ray(H, A)
m = Ray(K, B)
T = Intersect(l, m)
α = Angle(D, A, B)
β = Angle(E, B, A)
γ = Angle(A, D, B)
δ = Angle(B, E, A)
ε = Angle(B, E, K)
ζ = Angle(A, D, H)
η = Angle(E, B, K)
θ = Angle(H, A, D)
ι = Angle(T, A, B)
κ = Angle(T, B, A)