# What constitutes a direct proof?

Suppose I want to prove the following elementary theorem: The perpendicular bisector of the base of an isosceles triangle is the angular bisector of its vertex angle.

To do so, I would start by drawing the angular bisector of the vertex angle, and show that it must be a perpendicular bisector of the base. Since there exists a unique perpendicular bisector of the base, if we draw a perpendicular bisector of the base, we will inevitably draw what we just drew, the angular bisector of the vertex angle, hence proving the proposition.

This method of proof, which in Asia (and by Asia I mean Japan and Korea) is introduced as an example of an indirect proof (the translation of which I could not find, hence the question), is described as follows: Suppose I want to prove that B is A. I then prove that A is B instead, and then prove that B is unique. Thus this unique B must be A also, completing the proof.

So here are the questions: (1) Is the first proof given above (about the angular bisector of the vertex angle) an indirect proof? (2) Is there an English term for the method of proof described in the following paragraph?

• According to this reference there are two types of indirect proof: proof of the contrapositive and proof by contradiction. The proof you gave is neither of them.
– user
Commented Oct 18, 2023 at 13:42

A direct proof would start with the perpendicular bisector and then show that it bisects the angle.

Your proof is indirect - assuming the theorem false and deriving a contradiction. Implicitly, you say

Assume the perpendicular bisector is different from the angle bisector. Draw the angle bisector. "Inevitably" it is the same line, which is a contradiction.

In that not quite finished proof you have to demonstrate the inevitability, not just assert it.

• I thought that a proof is indirect if the original hypothesis is not assumed. In this case, the original hypothesis is that a line happens to be a perpendicular bisector of the base. It is definitely assumed; it equalling the angular bisector is the consequence of its uniqueness and the uniqueness of the angular bisector, and the fact that this unique angular bisector being a perpendicular bisector of the base? Commented Oct 17, 2023 at 18:51
• More precisely, the hypothesis here is: A line is a perpendicular bisector of the base of an isosceles triangle. Known theorems being employed are: (1) An angular bisector of the vertex angle is unique; (2) An angular bisector of the vertex angle perpendicularly bisects the base; (3) A perpendicular bisector of the base is unique. Hence this line that assumed to be a perp. bis. must be the unique one, which must also be the angular bisector. I think this is direct? Commented Oct 17, 2023 at 18:56
• If you have previously established the uniqueness of both bisectors then you could consider this a direct proof. Whatever you call it, I don't really like it. Presumably the arguments in the uniqueness proofs are just what you need to establish this theorem directly. I think you can often rephrase indirect proofs to sound/be direct in this way. Commented Oct 17, 2023 at 19:03
• This kind of proof is not uncommon in geometry - some theorems are most easily proven by using their converse. For example, the proof of the converse to the Pythagorean theorem: Suppose $a,b,c$ are lengths of a triangle satisfying $a^2 + b^2 = c^2$. Construct a right triangle with legs $a,b$, and then its hypotenuse must have length $c$. Then the triangles are congruent by SSS, so the original triangle must be a right triangle. I wouldn't call either of these proof by contradiction. I don't know what term if any would be correct. Perhaps "proof by uniqueness" or something like that. Commented Oct 17, 2023 at 19:10

1- direct proof:The triangle is isosceles. to prove the theorem. just drop a perpendicular from vertex A on the base BC. Mark it;s foot H. Right angle triangles ABH and ACH are equal due to common edge AH and equal hypotenuses which are equal sides AB and AC. This means $$\angle HAB=\angle HAC$$, that means AH is bisector of angle BAC. It Also means $$BH=CH$$ , this means AH is also the perpendicular bisector of base BC. We conclude that the perpendicular bisector of the base is also the bisector of the angle of the vertex A and vice versa.

2- Indirect proof: Suppose AH is not the bisector of angle BAC, then triangle ABH and ACH are not equal and $$\angle AHB \neq\angle AHC$$, also $$AH\neq CH$$, This contradicts the equality of segments CH and BH resulting from the fact that AH is the perpendicular bisector of base (said in in the statement).

• It is simpler as suggested by TS himself to drop at once the angle bisector because this does not introduce a new entity: a perpendicular from vertex $AB$ on the base $BC$. And the question was actually if it is (in English) direct or indirect proof.
– user
Commented Oct 18, 2023 at 13:32
• we can not argue using unproved 'facts' like inevitability in mathematics. We have to use known and proved facts or theorems. The argument of OP is neither direct nor indirect argument, unless is accepted as something new in math. Commented Oct 18, 2023 at 16:27