P

Q

R

Drag to change r

B

Bisecting an Angle

(Drag the blue points to change the angle)

First draw any circle, centered at P

Mark the 2 points where the circle intersects the two lines.
                      Label them P and Q.

Draw a circle with center P of radius

$ \left|PQ\right| $

Draw a circle with center P of radius

$ \left|PR\right| $

$ \left|PQ\right| = \left|PR\right| $

Call the intersection of the circles B

Draw the line through P and B

Claim:∠QPB = ∠RPB$ $

Consider 

△PRB

△PQB

and

We will show that they are congruent, using SSS.

$ \overline{PB} \cong \overline{PB} $

$ \overline{PQ} \cong \overline{PR}$

Since they are both radii of the circle with radius r

$ \overline{QB} \cong \overline{RB} $

So by SSS, the two triangles are congruent.

Since they are corresponding angles of two congruent triangles,
∠QPB = ∠RPB and we have bisected ∠QPR.

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A = (7.95818181818182, 4.305454545454545)
B = (1.5581818181818199, -2.1672727272727297)
a = Segment(A, B)
C = (13.121818181818181, -3.0945454545454574)
b = Segment(B, C)
D = (1.0672727272727291, 0.4872727272727256)
c = Circle(B, D)
E = Intersect(c, a)
F = Intersect(c, b)
d = Circle(F, B)
e = Circle(E, B)
G = Intersect(d, e, 2)
f = Line(B, G)
step = 1 (slider from 1 to 12, step 1)
g = Circle(B, E)
H = Intersect(g, b, 1)
I = Intersect(g, a, 1)
h = Circle(I, B)
k = Circle(H, B)
J = Intersect(h, k, 2)
α = Angle(A, B, J)
β = Angle(J, B, C)
K = (-0.14, 7.06)
L = (4.176363636363638, 6.632727272727274)
M = Intersect(g, a, 1)
N = Intersect(g, b, 1)
poly1, j, b_1, m = Polygon(B, M, J)
poly2, n, j_1, b_2 = Polygon(J, B, N)
i = Segment(B, J)
l = Segment(B, M)
p = Segment(B, N)
O = (-4.623636363636361, 5.76)
P = (0.23090909090909276, 7.178181818181819)