A way of directly proving the law of tangents without using circles.
Given:
1. Triangle ABC, with sides AB, AC, and opposite angles ACB, ABC.
2. Euclid.
3. The tangent of an angle is the ratio of the opposite side over the shorter adjacent side when that angle is part of a right triangle
To prove that the sum of two sides times the tangent of half of the difference between the angles opposite equals the difference between the two sides times the tangent of half of the sum of the angles opposite.
That is, that (AC-AB) x tangent((ABC+ACB)/2)=(AC+AB) x tangent((ABD-ACB)/2)
Take the two sides, such as may be AB and AC.
If the sides are equal
The difference between AB and BC is nothing.
In a triangle angles are to each other as their opposite sides.
Therefore the difference between angles ABC and ACB is nothing.
Nothing times something equals nothing.
According to the terms in the given, nothing equals nothing.
If the sides are not equal
Let AC>AB.
Cut off from AC line AD equal to AB.
Join BD
Cut BD in half at point E.
Draw AE through to point F on line BC.
AD was made equal to AB.
ED was cut equal to EB.
AE equals AE
Therefore traingles ADE and ABE are similar.
Therefore angles AEB and AED are equal.
Therefore line AF cuts line BD at right angles at E.
Any sums of all angles in a triangle are the same.
ABC+ACB+BAC=ABD+ADB+BAD
BAC is BAD
Therefore ABC+ACB=ABD+ADB
ABD=ADB
Therefore ABC+ACB=2ABD
Therefore ABD=(ABC+ACB)/2
ABC-ABD=ABC-(ABC+ACB)/2=(2ABC-(ABC+ACB))/2=(2ABC-ABC-ACB)/2=(ABC-ACB)/2
But ABC-ABD=EBF
Therefore EBF=(ABC-ACB)/2
AEB and BEF are right angles.
Triangles AEB and BEF are right triangles
Therefore tangent(ABE)=AE:EB
And tangent(EBF)=EF:BE
Cut DC in half at point G
Draw HGI parallel to FEA with point H on line BC
Draw AH parallel to BC
AB+AC:AC-AB :: AD+AC:AC-AD :: AD+AD+DC:AD+DC-AD :: 2AD+DC:DC
2AD+DC:DC :: AD+DC/2:DC/2 :: AD+DG:GC :: AG:GC
AI and HC are parallel
Therefore AG:GC :: IG:GH
But DG=half DC
And DE=half DB
Therefore EG is parallel to BC
Therefore AI, EG, FH are parallel
And IH, AF are parallel
Therefore IG:GH=AE:EH
With all these discovered we can say:
Tangent((ABC+ABD)/2):tangent((ABC-ABD)/2)=tangent(ABE):tangent(EBF) :: (AE:EB):(EF:EB) :: AE:EF
And AE:EF=IG:GH :: AG:GC :: AB+AC:AC-AB.
Therefore tangent((ABC+ABD)/2):tangent((ABC-ABD)/2) :: AB+AC:AC-AB
Therefore tangent((ABC+ABD)/2)(AC-AB)=tangent((ABC-ABD)/2)(AB+AC)
Which is what was to be proven. Q.E.D.
