Solution to puzzle 22 generalization

We will use the law of sines, on triangles BED and BCD, as in the original puzzle, and on triangle BEC.

We have BAC = 2a, DBC = b, and ECB = c.  Let EDB = x.

Applying the law of sines to:
BED,  BE/sinx = BD/sin(90°+a+b−x),
BCD,  BC/sin(90°+a−b) = BD/sin(90°−a),
BEC,  BE/sinc = BC/sin(90°+a−c).

Therefore BD = BE · cos(a+b−x) / sinx = BC · cosa / (cos(a−b)

Also BC = BE · cos(a−c) / sinc

Therefore cos(a+b−x) / sinx = cos(a−c) cos a / cos(a−b) sin c

Using trigonometric identity cos(y−z) = cosy cosz + siny sinz,
cos(a+b) cot x + sin(a+b) = cos(a−c) cos a / cos(a−b) sin c.

Using trigonometric identity 2 cos y cos z = cos(y−z) + cos(y+z):

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Remarks

The table below shows all solutions for which all angles are integers (when measured in degrees), and for which DBC is greater than ECB.

Integer degree solutions

BAC	DBC	ECB	EDB
4	46	4	2
4	46	44	42
8	47	8	4
8	47	43	39
12	42	18	12
12	42	30	24
12	48	12	6
12	48	42	36
12	57	33	15
12	57	42	24
12	66	42	12
12	66	54	24
12	69	21	3
12	69	66	48
12	72	42	6
12	72	66	30
16	49	16	8
16	49	41	33
20	50	20	10
20	50	40	30
20	60	30	10
20	60	50	30
20	65	25	5
20	65	60	40
20	70	50	10
20	70	60	20
24	51	24	12
24	51	39	27
28	52	28	14
28	52	38	24
32	53	32	16
32	53	37	21
36	54	36	18
40	55	35	15
40	55	40	20
44	56	34	12
44	56	44	22
48	57	33	9
48	57	48	24
52	58	32	6
52	58	52	26
56	59	31	3
56	59	56	28
72	39	21	12
72	39	27	18
72	42	24	12
72	42	30	18
72	48	24	6
72	48	42	24
72	51	39	9
72	51	42	12
120	24	12	6
120	24	18	12

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