Presented here are 8 distinct proofs that in the following diagram, a+b=c:

(Throughout these proofs the above naming conventions will be used.)
The distinct proofs are:
From the diagram, a = tan -1(1/3), b = tan -1(1/2), and c = tan -11.
We have to prove:
| a + b | = | c | |
| or | tan -1(1/3) + tan -1(1/2) | = | tan -11 |
| or | tan[tan -1(1/3) + tan -1(1/2)] | = | 1. |
Using tan double angle formula, tan( x+y) = [tan( x)+tan( y)]/[1-tan( x)tan( y)], we may rewrite the equation as:
| tan[tan -1(1/3) + tan -1(1/2)] | = | ![]() |
| = | [1/3 + 1/2]/[1-(1/3)(1/2)] | |
| = | (5/6)/(5/6) | |
| = | 1 |
Hence the result is proved.
From diagram:

sin( a+b) = sin( c), so a+b=c. Hence the result is proved.
By altering the diagram:

It can be seen that x=3+2=5.
d can be found using the cosine rule ( a 2+ b 2-2 ab cos C = c 2), so

| So | a + b | = | 180° - d |
| = | 180° - 135° | ||
| = | 45° | ||
| c = | tan -11 | = | 45° |
| So | a + b | = | c. |
Hence the result is proved.
![]() |
![]() |

| c = | tan -11 | = | 45° |
| So | a + b | = | c. |
Hence the result is proved.
Let d = ( x,y) be any point on the xy-plane.
Let d 1 be d rotated a degrees around the origin:

Let d 2 be d 1 rotated b degrees around the origin: (or another way: d rotated ( a+b) degrees around the origin)

Let d 3 be d rotated c degrees around the origin:

So d 2 = d 3 .
Hence a rotation by a+b is the same as a rotation by c degrees. Hence a+b=c (as none of the angles are greater than 90°).

z = x +iy = r exp( i tan -1 y/x)
Let z´ = z rotated 2( a + b) degrees around origin.
| So | z´ | = r exp( i tan -1 y/x + 2 i( a + b)) |
| = r [cos(tan -1 y/x + 2( a + b)) + i sin(tan -1 y/x + 2( a + b))] | ||
| = r [cos(tan -1 y/x)cos(2( a + b)) - sin(tan -1 y/x)sin(2( a + b)) + i(sin(tan -1 y/x)cos(2( a + b)) + cos(tan -1 y/x)sin(2( a + b)))] |
Since cos(2( a + b)) = 2 cos 2( a + b) - 1
| and | cos( a + b) | = cos a cos b - sin a sin b |
![]() |
so we have cos(2( a + b)) = 0.
Similarly sin(2( a + b)) = 2sin( a + b)cos( a + b)
| and | sin( a + b) | = sin a cos b + cos a sin b |
![]() |
so we have sin(2( a + b)) = 1.
| Therefore | z´ | = r [0 - sin(tan -1 y/x) + i(0 + cos(tan -1 y/x))] |
| = r [ i cos(tan -1 y/x) - sin(tan -1 y/x)] |
![]() |
|
Let z´´ = z rotated by 90° around the origin. (Multiplying a complex number by i is equivalent to a 90° rotation.)
z´´ = i(x + iy) = -y + ix.
So a rotation of 2( a + b) is equal to a rotation of 90°, hence
| 2( a + b) | = 90° |
| a + b | = 45° |

By looking at the leftmost unit square this diagram can be drawn:
![]() |
|
| a + b = c | <=> | x + x + y = x + y + z, |
| <=> | x = z |
Hence it must be proven that x = z:
![]() |
|
|
![]() |
Hence ADF and ABC are similar triangles and x=z => a + b = c.

Let B be the point ( x B,y B) on the line y = -(1/2) x, a distance of 1 from the origin.

The line perpendicular to y = -(1/2) x at B has equation:

This line intersects y = (1/3) x at A = ( x A,y A).
