圖解算幾不等式
Author: 藍邦偉  老師

$ 圖解算幾不等式 $
$\;\;\\;\;\frac{a+b}{2}≥\sqrt{ab}$
$\;\;\;拉動A，\;B，\;C\;觀察$

$ a $

$ b $

$ \frac{a+b}{2} $

$ \sqrt{ab} $

$A  $

$ B $

$ H $

$ M $

$ C $

1

$ 顯示說明 $

u = Segment((x(Corner[1]), 0), (x(Corner[3]), 0))
v = Vector((0, y(Corner[1])), (0, y(Corner[3])))
O = (0, 0)
Q_2: 0.75 * Corner[1] + 0.25 * Corner[2]
Q_3: (x(Q_2), y(Corner[4]))
poly1, q_1, q_2, q_4, q_3 = Polygon(Corner[1], Q_2, Q_3, Corner[4])
A = Point(u)
B = Point(u)
c = Semicircle(A, B)
C = Point(c)
M = Midpoint(B, A)
H: (x(C), 0)
a = Segment(C, M)
b = Segment(C, H)
d = Segment(C, A)
e = Segment(C, B)
f = LineBisector(A, H)
g = LineBisector(H, B)
E = (6.334749161568283, 3.3075312569308806)
h = Line(E, u)
F = Intersect(h, f)
G = Intersect(g, h)
k = CircleArc(F, A, H)
p = CircleArc(G, H, B)
i = LineBisector(C, H)
j = LineBisector(C, M)
H_1 = Point(i)
q = CircleArc(H_1, C, H)
I = Point(j)
r = CircleArc(I, M, C)
α = Angle(B, H, C)
t1 = false
text1: "$ a=\overline{AH}=" + (LaTeX[Distance[A, H]]) + " $
$b=\overline{BH}=" + (LaTeX[Distance[B, H]]) + "$
$\frac{a+b}{2}=\overline{MA}=\overline{MB}$
$\;\;\;\;\;\;\;\;\;=\overline{MC}=" + (LaTeX[Distance[M, A]]) + "$
$\sqrt{ab}=\overline{CH}=" + (LaTeX[Distance[H, C]]) + "$"