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):

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
BACDBCECBEDB
44642
4464442
84784
8474339
12421812
12423024
1248126
12484236
12573315
12574224
12664212
12665424
1269213
12696648
1272426
12726630
1649168
16494133
20502010
20504030
20603010
20605030
2065255
20656040
20705010
20706020
24512412
24513927
28522814
28523824
32533216
32533721
36543618
40553515
40554020
44563412
44564422
4857339
48574824
5258326
52585226
5659313
56595628
72392112
72392718
72422412
72423018
7248246
72484224
7251399
72514212
12024126
120241812