α

h

Höhensatz des Euklid

Beweis wegen

\frac{h}{p}=\frac{q}{h}

A = Point(xAxis)
B = Point(yAxis)
c = Segment(A, B)
c_2 = Semicircle(A, B)
C = Point(c_2)
g = OrthogonalLine(C, c)
f_2 = Line(B, g)
D = Intersect(g, c)
p = Segment(A, D)
p_1 = Circle(D, p)
P, Q = Intersect(p_1, g)
s_1 = CircleArc(D, P, B)
S = Intersect(p_1, s_1)
t_1 = OrthogonalLine(S, g)
T = Intersect(t_1, f_2)
E = Point(g)
d1, c_1, a, b = Polygon(A, B, C)
γ = Angle(A, C, B)
h = Segment(C, D)
i = Segment(D, E)
q = Segment(D, B)
α = Angle(B, A, C)
β = Angle(D, C, B)
Vieleck1, f, j, k, l, F, G = Polygon(C, B, 4)
Vieleck2, m, n, r, s, H, I = Polygon(A, C, 4)
Vieleck3, t, d, e, f_1, J, K = Polygon(B, A, 4)
L, M = Intersect(g, Vieleck3)
v1, j_1, l_1, d_1, a_1 = Polygon(J, L, D, A)
v2, l_2, k_1, b_1, d_2 = Polygon(L, K, B, D)
e_1 = Circle(A, c)
g_1 = CircleArc(A, J, B)
Vieleck4, h_1, i_1, m_1, n_1, N, O = Polygon(D, C, 4)
q_1 = Line(A, g)
r_1 = OrthogonalLine(P, g)
R = Intersect(q_1, r_1)
v3, g_2, h_2, i_2, j_2 = Polygon(D, B, T, S)
e_2 = CircleArc(D, A, S)
U = (-10.381688888888888, 2.0802666666666676)
V = (-8.681688888888889, 2.0802666666666676)
Text2: "h^2=" + (LaTeX[Vieleck4]) + ""
Text3: "p\cdot q=" + (LaTeX[Vieleck4]) + ""