# proof of the fact that the set of points equidistant from sides of an angle form a bisector of the angle

Lets have a line divided into two parts by a point S. Lets construct a ray from S that has an angle of alpha with the left ray beginning with S. Lets construct another angle of measure alpha, this time using the right ray as a side of our angle. If alpha is arbitrary, then the angle between them is also arbitrary. Lets construct a line perpendicular to line on which S lies that goes through S. Lets choose an arbitrary point on the line P (the point should be above S). Lets construct two lines going through P that are perpendicular to the sides of angles we have already constructed. Lets name the intersection points X and Y. I claim that when we draw a line segment between X and Y we get a segment perpendicular to our base line (the one with S). I, further claim that the triangle PXY is isoceles. (We use the fact that the left our two identical angles is congruent to its alternate interior angle. We also know that the PXY is equal to 90-alpha degrees We can use the same deduction to show that PYX is 90-alpha degrees.) Therefore the distance between PX and PY is identical.

Is the proof correct?

• The construction is a little hard to follow. (BTW, each "Lets" should be "Let's".) Even so, once the construction sets the stage, the proof itself consists of your your two claims and the final conclusion. However, the first claim goes without justification, and it's not immediately clear how the second claim's justification actually applies. This is an incomplete proof. – Blue Nov 22 '13 at 16:47
• By the construction being hard to follow, do you mean that you were unable to retrace its steps? As for the rest of your comment, I will try to fix that. – Adam Nov 22 '13 at 16:52
• "Left" and "right" are problematic adjectives in a diagram-less description. (What if I'd drawn my original line vertically?) Also, the phrase "[the] line on which $S$ lies" is ambiguous; point $S$ is on multiple lines by that time. Stuff like that. – Blue Nov 22 '13 at 17:08
• Here's an attempt at reformulation (although, really, a digram would be best): Consider collinear (& distinct) points $A$, $B$, $S$, with $S$ between $A$ and $B$. Let point $P$ be such that $\overleftrightarrow{PS}\perp\overleftrightarrow{AB}$. For a given point $X$ (not on $\overleftrightarrow{AB}$), we can find $Y$ (on the same side of $\overleftrightarrow{AB}$) such that $\angle XSA \cong \angle YSB$; we may take $X$ and $Y$ specifically to be the feet of perpendiculars from $P$ to $\overleftrightarrow{SX}$ and $\overleftrightarrow{SY}$. – Blue Nov 22 '13 at 17:09
• Here's a cleaner setup to achieve your goal: "Consider angle $\angle XSY$ with interior ray $\overrightarrow{SP}$, where we may take $X$ and $Y$ specifically to be the feet of perpendiculars from $P$ to the sides of the angle." Done. Now, show that (1) assuming $\angle PSX \cong\angle PSY$, we get $\overline{PX}\cong\overline{PY}$; and, conversely, (2) assuming $\overline{PX}\cong\overline{PY}$, we get $\angle PSX \cong \angle PSY$. Each direction involves a straightforward triangle congruence argument. (Your initial line including $S$ is an unnecessary distraction.) – Blue Nov 22 '13 at 17:18