Prove all right angles are congruent.
I only have to prove one side to this argument, so I just need to the the other argument.
So basically, if two angles are right, then they must be congruent is what I am trying to prove.
All I have is my assumption that the two angles are right. And conclusion, therefore the angles are congruent.
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$\begingroup$We say that the angle $\measuredangle AOB$ is the supplement of the angle $\measuredangle Y$ if the latter is congruent to an adjacent angle $\measuredangle BOC$ to $\measuredangle AOB$ such that the points $A$, $O$ and $C$ are colineal.
Using this definition and the fact that an angle is Right iff it's congruent to one of its supplements (by definition), you can prove that all right angles are congruent as follows:
Let $\measuredangle AOB$ be a right angle, then it's congruent to one of its supplements (and therefore to all of them). Let $\measuredangle BOC$ be an adjacent supplement of $\measuredangle AOB$, then $\measuredangle AOB \cong \measuredangle BOC$ and $A$, $0$, $C$ are colineal.
Now let $Y$ be any other right angle and consider $D$ an exterior point of $\measuredangle AOB$ such that $\measuredangle AOD$ is a right angle congruent to $Y$. Here you have to prove that: $B$, $O$ and $D$ are colineal and once you have this prove that $\measuredangle AOB\cong \measuredangle AOD$ (using "vertex opposites" arguments), I'm leaving that to you.
$\endgroup$ $\begingroup$If you define a right angle as the bisector of a straight angle - analogous to the idea of folding over a straight edge to get a fold perpendicular to that fold, with the same angle either side of the fold line to the straight line - it is clear that since straight lines are taken as similar, and you can create a straight angle by placing a point on a straight line, all bisectors of straight angles - all right angles - are also congruent.
$\endgroup$ $\begingroup$A straight angle has two right angles. If two angles are such that a supplement of the one equals itself then each must be a right angle. Since only a single parameter is involved congruence is established.
$\endgroup$ $\begingroup$By supplements do you mean they sum to 180 degrees or would form a line if they were adjacent? Have you considered a proof by contradiction? I love those! Then you could assume you have two right angles which are not congruent. Going that route we are given angle A is a right angle, angle 1 is a right angle, and angle A is not congruent to angle 1. Then there exists angle B which is a congruent supplement and linear pair to angle A and angle 2 which is a congruent supplement and linear pair to angle 1. Assuming that all lines are the same degrees (180 or whatever), angle B and angle 2 are both half lines (90 degrees or whatever). So angle B and angle 2 have the same measure and are congruent, which should bring about our contradiction. Angle A is congruent to angle B, angle B is congruent to angle 2, angle 2 is congruent to angle 1, so angle A is congruent to angle 1, except that it isn't by definition. Might have to do a little cleaning up in the middle there if you move from congruence to numbers and back.
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