Theorems concerning quadrilateral properties
So we have a parallelogram right over here. And what I want to prove is that its diagonals bisect each other. So the first thing that we can think about-- these aren't just diagonals. These are lines that are intersecting, parallel lines. So you can also view them as transversals. And if we focus on DB right over here, we see that it intersects DC and AB. And since we know that they're parallel-- this is a parallelogram-- we know the alternate interior angles must be congruent. So that angle must be equal to that angle there. And let me make a label here. Let me call that middle point E. So we know that angle ABE must be congruent to angle CDE by alternate interior angles of a transversal intersecting parallel lines. Now, if we look at diagonal AC-- or we should call it transversal AC-- we can make the same argument. It intersects here and here. These two lines are parallel. So alternate interior angles must be congruent. So angle DEC must be-- so let me write this down-- angle DEC must be congruent to angle BAE, for the exact same reason. Now we have something interesting, if we look at this top triangle over here and this bottom triangle. We have one set of corresponding angles that are congruent. We have a side in between that's going to be congruent. Actually, let me write that down explicitly. We know-- and we proved this to ourselves in the previous video-- that parallelograms-- not only are opposite sides parallel, they are also congruent. So we know from the previous video that that side is equal to that side. So let me go back to what I was saying. We have two sets of corresponding angles that are congruent, we have a side in between that's congruent, and then we have another set of corresponding angles that are congruent. So we know that this triangle is congruent to that triangle by angle-side-angle. So we know that triangle-- I'm going to go from the blue to the orange to the last one-- triangle ABE is congruent to triangle-- blue, orange, then the last one-- CDE, by angle-side-angle congruency. Now, what does that do for us? Well, we know if two triangles are congruent, all of their corresponding features, especially all of their corresponding sides, are congruent. So we know that side EC corresponds to side EA. Or I could say side AE corresponds to side CE. They're corresponding sides of congruent triangles, so their measures or their lengths must be the same. So AE must be equal to CE. Let me put two slashes since I already used one slash over here. Now, by the same exact logic, we know that DE-- let me focus on this-- we know that BE must be equal to DE. Once again, they're corresponding sides of two congruent triangles, so they must have the same length. So this is corresponding sides of congruent triangles. So BE is equal to DE. And we've done our proof. We've shown that, look, diagonal DB is splitting AC into two segments of equal length and vice versa. AC is splitting DB into two segments of equal length. So they are bisecting each other. Now let's go the other way around. Let's prove to ourselves that if we have two diagonals of a quadrilateral that are bisecting each other, that we are dealing with a parallelogram. So let me see. So we're going to assume that the two diagonals are bisecting each other. So we're assuming that that is equal to that and that that right over there is equal to that. Given that, we want to prove that this is a parallelogram. And to do that, we just have to remind ourselves that this angle is going to be equal to that angle-- it's one of the first things we learned-- because they are vertical angles. So let me write this down. Let me label this point. Angle CED is going to be equal to-- or is congruent to-- angle BEA. Well, that shows us that these two triangles are congruent because we have our corresponding sides that are congruent, an angle in between, and then another side. So we now know that triangle-- I'll keep this in yellow-- triangle AEB is congruent to triangle DEC by side-angle-side congruency, by SAS congruent triangles. Fair enough. Now, if we know that two triangles are congruent, we know that all of the corresponding sides and angles are congruent. So for example, we know that angle CDE is going to be congruent to angle BAE. And this is just corresponding angles of congruent triangles. And now we have this transversal of these two lines that could be parallel, if the alternate interior angles are congruent. And we see that they are. These two are kind of candidate alternate interior angles, and they are congruent. So AB must be parallel to CD. Actually, I'll just draw one arrow. AB is parallel to CD by alternate interior angles congruent of parallel lines. I'm just writing in some shorthand. Forgive the cryptic nature of it. I'm saying it out. And so we can then do the exact same-- we've just shown that these two sides are parallel. We could then do the exact same logic to show that these two sides are parallel. And I won't necessarily write it all out, but it's the exact same proof to show that these two. So first of all, we know that this angle is congruent to that angle right over there. Actually, let me write it out. So we know that angle AEC is congruent to angle DEB. They are vertical angles. And that was our reason up here, as well. And then we see the triangle AEC must be congruent to triangle DEB by side-angle-side. So then we have triangle AEC must be congruent to triangle DEB by SAS congruency. Then we know that corresponding angles must be congruent. So for example, angle CAE must be congruent to angle BDE. And this is they're corresponding angles of congruent triangles. So CAE-- let me do this in a new color-- must be congruent to BDE. And now we have a transversal. The alternate interior angles are congruent. So the two lines that the transversal is intersecting must be parallel. So this must be parallel to that. So then we have AC must be parallel to be BD by alternate interior angles. And we're done. We've just proven that if the diagonals bisect each other, if we start that as a given, then we end at a point where we say, hey, the opposite sides of this quadrilateral must be parallel, or that ABCD is a parallelogram.