 My name is Adam. I'm a student of Dr. Missildine's at Southern Utah University and today we're going to be exploring the midpoint theorem Which says that every segment has a unique midpoint and this is in Hilbert's congruence geometry Which has Hilbert's axioms of incidents betweenness and congruence so to start We're going to just Have a segment BC I'm going to say that the point a Is a point off of BC over here All right By line determination, there's a line between a and c and by extension we can extend that out To another point D All right Then we can finish this triangle over here by connecting b and a and we see that the angle dcb is an exterior angle of the triangle CAB and most specifically to the remote angle b out here So this angle Must be smaller than angle dcb by the exterior angle theorem And it must therefore by by angle translation be congruent to an angle inside the half planes created by BCd and We'll call this point E and without loss of generality We're going to go ahead and choose a point E on this ray CE such that CE and a B are congruent Okay, and notice Since E is on the same side of BC as D This is on the inside the angle BCD and a is on the opposite side of D of BC as D Then by plane separation E is on the opposite side of BC as a and additionally Since by the alternate interior angle theorem a B and CE are parallel I'll go ahead and write that and the point E must therefore be on the inside of the angle BAC as we show here and That lets us use the crossbar theorem because this segment BC Since E is on the inside of it of the angle BAC And it must the ray AE must intersect BC at some point with that crossbar theorem we referenced And we're going to go ahead and call that point of intersection F. All right Now we don't have it as a midpoint yet, but let's go ahead and do some triangle congruences to get that Okay, well, we know that the angle BFA is congruent to the angle EFC Because they're vertical angles, so we'll show that and then by the triangle congruence angle angle side because AFB FBA and the segment BA are congruent to EFC FCE and the segment CE and the triangle ABF is congruent to the triangle CEF And as corresponding parts of congruent triangles are congruent that gives us These segments here that we're looking for BF and FC are congruent So by definition F is the midpoint of B and C So what that's done so far is given us a midpoint so we can guarantee a midpoint But we need still need to prove that there's only one and it's a unique midpoint. So let's switch gears over here Look at another piece We'll use we use primes over here so a primes B or so we'll see prime The segment a prime C prime over here. Let's say That way of contradiction That it has two points That are both the midpoint of a C prime a prime C prime. I want to call those points B and B prime All right, and these aren't related to be over here. This is a different Different example All right, so by definition since B is the midpoint of a prime C prime Then a B is congruent to BC and since B prime is also the midpoint of a prime C prime Then a B prime or a prime B prime is congruent to B prime C prime All right So if we set up some inequalities, we'll know that a prime B prime We know that a prime B prime the segment is bigger Without loss of generality at least greater than or equal the segment a B Which is congruent to the segment BC prime which is Greater than or equal to the segment B prime C prime Which is congruent to a prime B prime Let's all these inequalities Have to be congruences because a prime B prime is obviously congruent to a prime B prime So that means that a prime B prime is congruent to the segment a prime B But since they're both in the same ray And by uniqueness of segment translation That means that B equals B prime Therefore B is the only unique midpoint of a prime C prime So we proved over here That midpoints exist. There's at least one midpoint in every segment and we proved over here Now there's no more than one unique midpoint of any segment And that is the midpoint theorem. Thank you for watching