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Proof: Rhombus diagonals are perpendicular bisectors

Sal proves that the diagonals of a rhombus are perpendicular, and that they intersect at the midpoints of both. Created by Sal Khan.

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Video transcript

We're told that quadrilateral ABCD is a rhombus. And what they want us to prove is that their diagonals are perpendicular, that AC is perpendicular to BD. Now let's think about everything we know about a rhombus. First of all, a rhombus is a special case of a parallelogram. In a parallelogram, the opposite sides are parallel. So that side is parallel to that side. These two sides are parallel. And in a rhombus, not only are the opposite sides parallel-- it's a parallelogram-- but also, all of the sides have equal length. So this side is equal to this side, which is equal to that side, which is equal to that side right over there. Now, there's other interesting things we know about the diagonals of a parallelogram, which we know all rhombi are parallelograms. The other way around is not necessarily true. We know that for any parallelogram-- and a rhombus is a parallelogram-- that the diagonals bisect each other. So for example, let me label this point in the center. Let me label it point E. We know that AE is going to be equal to EC. I'll put two slashes right over there. And we also know that EB is going to be equal to ED. So this is all of what we know when someone just says that ABCD is a rhombus, based on other things that we've proven to ourselves. Now we need to prove that AC is perpendicular to BD. So an interesting way to prove it-- and you can look at it just by eyeballing it-- is if we can show that this triangle is congruent to this triangle and that these two angles right over here correspond to each other, then they have to be the same. And they'll be supplementary, and then they'll be 90 degrees. So let's just prove it to ourselves. So the first thing we see is we have a side, a side, and a side; a side, a side, and a side. So we can see that triangle-- let me write it here. Let me [? do this ?] in a new color. We see that triangle ABE is congruent to triangle CBE. And we know that by side-side-side congruency. We have a side, a side, and a side; a side, a side, and a side. And then once we know that, we know that all the corresponding angles are congruent. And in particular, we know that angle AEB is going to be congruent to angle CEB. Because they are corresponding angles of congruent triangles. So this angle right over here is going to be equal to that angle over there. And we also know that they are supplementary. Let me write it this way. They're congruent, and they are supplementary. These two are going to have the same measure, and they need to add up to 180 degrees. So if I have two things that are the same thing and they add up to 180 degrees, what does that tell me? So that tells me that the measure of angle AEB is equal to the measure of angle CEB, which must be equal to 90 degrees. They're the same measure, and they are supplementary. So this is a right angle, and then this is a right angle. And obviously, if this is a right angle, this angle down here is a vertical angle. That's going to be a right angle. If this is a right angle, this over here is going to be a vertical angle. And you see the diagonals intersect at a 90-degree angle. So we've just proved-- so this is interesting. A parallelogram, the diagonals bisect each other. For a rhombus, where all the sides are equal, we've shown that not only do they bisect each other but they're perpendicular bisectors of each other.