$t=%v$

畢氏定理

$Step=%v$

a

b

c

直角三角形ABC中，\;若a=\overline{BC},\;b=\overline{CA},\;c=\overline{AB},\;則:\;a²+b²=c²

將4個兩股為a,\;b的直角三角形，\;圍成一個邊長為\;a+b\;的正方形.

a

b

b

a

b

a^2+b^2=c^2

c²

a²

b²

A = (3.56, -1.82)
B: A + (6, 0)
C: B + (0, 6)
D: A + (0, 6)
a = Segment(A, B)
P = Point(a)
M: 0.5 * A + 0.5 * C
Q = Rotate(P, 1.5707963267948966, M)
R = Rotate(Q, 1.5707963267948966, M)
S = Rotate(P, -1.5707963267948966, M)
b = Segment(B, C)
c = Segment(D, C)
d = Segment(D, A)
t = 0 (slider from 0 to 1, step 0.01)
C' = Translate(C, Vector[t * Vector[R, S]])
Q' = Translate(Q, Vector[t * Vector[R, S]])
R' = Translate(R, Vector[t * Vector[R, S]])
list1: {Polygon[C', Q', R']}
D' = Translate(D, Vector[t * Vector[R, C]])
S' = Translate(S, Vector[t * Vector[R, C]])
R'_1 = Translate(R, Vector[t * Vector[R, C]])
list2: {Polygon[D', S', R'_1]}
B' = Translate(B, Vector[t * Vector[Q, C]])
P' = Translate(P, Vector[t * Vector[Q, C]])
Q'_1 = Translate(Q, Vector[t * Vector[Q, C]])
list3: {Polygon[B', P', Q'_1]}
List_A: {{Polygon[C', Q', R']}, {Polygon[D', S', R'_1]}, {Polygon[B', P', Q'_1]}}
listB: {Polygon[Translate[C, Vector[t * Vector[R, S]]], Translate[R, Vector[t * Vector[R, S]]], Translate[Q, Vector[t * Vector[R, S]]]], Polygon[A, P, S]}
listC: {Polygon[Translate[D, Vector[t * Vector[R, C]]], Translate[R, Vector[t * Vector[R, C]]], Translate[S, Vector[t * Vector[R, C]]]], Polygon[Translate[B, Vector[t * Vector[Q, C]]], Translate[P, Vector[t * Vector[Q, C]]], Translate[Q, Vector[t * Vector[Q, C]]]]}
poly1, a_1, b_1, c_1, d_1 = Polygon(A, B, C, D)
s = 0 (slider from 0 to 3, step 1)
Poly1, p, a_2, s_1 = Polygon(A, S, P)
α = Angle(P, A, S)
text3: "" + (LaTeX[ta]) + ""