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## High school geometry

### Course: High school geometry > Unit 3

Lesson 7: Proofs of general theorems# Geometry proof problem: squared circle

CCSS.Math:

Sal finds a missing angle using triangle congruence in a diagram that contains sector of a circle inscribed within a square. Created by Sal Khan.

## Want to join the conversation?

- How was it proven that Line EC and Line EB and Line BC are all equal?(4 votes)
- Because ABCD is a square, AB=BC=CD=DA. Because arc AC is part of circle B, that means BE is a radius as well as BA and BC and are, therefore, all equal. When Sal drew EC, he created triangle ECG and showed it was congruent to triangle EBG by SAS. Since BE and EC are both the hypotenuse of congruent triangles, they are equal. So EC=EB=BC.(5 votes)

- So its still considered a triangle if one or more of its legs are not a straight line? As long as it still has three legs that total to 180 degrees?(2 votes)
- Nope. A triangle must have 3 straight lines with interior angles adding up to 180 degrees.

Happy to help!(9 votes)

- He said that the harder geometry problems revolve around drawing the right triangles and visualizing the right lines. How can I get there(Knowing which lines/triangles to draw)?(4 votes)
- In respect to this, a good way to train your brain is just by practicing different types of problems altogether. On KA, a good way to do this is by taking the course challenge multiple times. Of course, you will also need to know the material well enough to use it when it's needed.

You might find it helpful to work on a difficult problem backwards. Say, if you have four lines that intersect at random angles and you need to find one specific measurement that is not offered, you might work backwards from the angle trying to figure out how to get everything you need. You could also work it forwards, finding*all*the measurements until you know how to get what you need.

But these are the difficulties of math: finding what operations you need to perform and not making small (or large!) mistakes. Anyone else who has tips is welcome to reply, as well. This answer to this question might very well be different to everyone!(3 votes)

- At7:15, How can you just add x+x+30=180 degrees to solve for the missing angles? How would you which one would be 75 degrees?(3 votes)
- They are both 75 because it is an isosceles triangle (two of its sides and two of its interior angles are the same)(4 votes)

- How often will we use proof theorems/problems in real life?(3 votes)
- The math itself may not be too useful, but the critical thinking and logic is.(3 votes)

- At6:52wouldn't it make sense for the missing angle to be 60˚, because CE could act as a transversal to ∆DEC, so angle CEB = angle DCE. So then the equation would be 2x + 60 = 180. So the two base angles for ∆DEC would be 60˚ right? Wouldn't that be true? Because most of these diagrams aren't draw to scale.(2 votes)
- Would the triangles at2:59be using the postulate RSH?(2 votes)
- At2:07we draw the segment EC. So, my question is can you add any segments you want in the square as we did for EC to solve a problem? Or are there any rules for adding such segments?(1 vote)
- No, you can add whatever you want! There are no rules as to what you can add to a problem(2 votes)

- At4:42I did something different, why is this incorrect?

Instead of doing Sal's solution, I removed the curved line of EC. This left a triangle, that was SS congruent (therefore not congruent) with the former. I.e ∆DEC SS ∆BEC, from DC = BC and EC = EC.

From there, I figured, that since it is a square, the sides are parallel. We can see, that FG and DC are parallel, since FG is perpendicular (a flat line) and DC is a flat side of a square.This means, that BC is a transversal, which means, that ∆DEC must at least share one angle with ∆BEC, and they're therefore both kongruent, which means I got the result, that BED = 120.

Why is this incorrect?(1 vote)- Several issues in the beginning you say SS congruent (therefore not congruent) which appears to conflict with each other, so I have no clue what you are talking about there. Secondly, there is no such thing as SS congruence, it has to be SSS congruence and it appears to be obvious that BE is not equal to DE.(2 votes)

- Could you make Tri EBG and Tri ECG congruent by using Hypotenuse leg ?(1 vote)

## Video transcript

So we're told that quadrilateral
ABCD is a square, which tells us that the four
sides have equal length and that all the interior
angles are 90 degrees. We also know that FG is a
perpendicular bisector of BC. So we've already shown
that it's perpendicular, that this is a 90-degree angle. But it also bisects
BC, so this length is equal to that length
right over there. Then they tell us that arc AC--
that's a curve on top-- arc AC is part of circle B. So this
is a circle centered at B. So if this is the
center of the circle, this is part of that circle. It's really kind of the bottom
left quarter of that circle. And then given that
information, they want us to find what the
measure of angle BED is. So what is BED? So we need to figure
out the measure of this angle right over here. And I encourage you to
pause it and try it out. And you might
imagine-- well, you could pause it and try
it without any hints. And now I will give you a hint
if you tried it the first time and you weren't able to do it. And you should pause again
after this hint-- try to draw some triangles
that maybe split up this angle into a couple
of different angles. And it might be a
little bit easier. You might be able to use some
of what we know about triangles. Now, with that said, I
will try to solve it. And you should
pause at any point where you think that you know
exactly how to do this and try to do it yourself. So the trick here to realize
that this is a circle. And so any line that goes
between B and any point on this arc is going to be equal
to the radius of the circle. So AB is equal to the
radius of the circle. BE is equal to the
radius of the circle. And we can keep drawing
other things that are equal to the
radius of the circle. BC is equal to the
radius of the circle. So let's think about
it a little bit. If we were to draw it-- and
a lot of trickier geometry problems really all revolve
around drawing the right lines or visualizing the
right triangles. And I'll do one right here that
might open up a lot for you in terms of thinking about
how to do this problem. So let me draw segment EC. I'll draw that as
straight as possible. I can draw a better job
of that, so segment EC. Now something
becomes interesting. Because what is the relationship
between triangle EBG and triangle ECG? Well, they both definitely
share this side right over here. They both share side EG. And then BG is equal
to GC, and they both have 90-degree angles. You have a 90-degree angle here. You have a 90-degree
angle there. So you see by side-angle-side
that these two triangles are going to be congruent. So we know that triangle EBG is
congruent to triangle ECG-- I should emphasize
the C, not the E-- by side-angle-side congruency. And that also tells us that
all of the corresponding angles and sides are going
to be the same. So that tells us right there
that EC is equal to EB. So we know that
EB is equal to EC. And what also is
equal to that length? Well, once again, this is
the radius of the circle. BE is one radius of the
circle going from the center to the arc. But so is BC. It is also a radius of the
circle going from the center to the arc. So this is also equal to BC. So I could draw the
three things right here. I'm referring to the
entire thing, not just one of the segments, all of BC. So what kind of a
triangle is this right over here, triangle BEC? Triangle BEC is equilateral. And we know that because
all three sides are equal, so that tells us that all
of its angles are equal. So that tells us that
the measure of angle BEC-- we're not done yet, but it
gets us close-- is 60 degrees. So the measure of angle BEC
right over there is 60 degrees. So that gives us
part of the problem. BEC is part of the angle BED. If we can just figure out
the measure of angle CED now, if we can figure out this
angle right over here, we just add that to 60
degrees, and we're done. We've figured out
the entire BED. Now, let's think about how we
can do this right over here. So there's a couple
of interesting things that we already do know. We know that this
right over here is equal to the
radius of the circle. And we also know that this
length down here-- this is a square-- we know that
this length down here is the same as this length
up here, that these are the exact same length. And this is equal to the
radius of the circle. We already put these
three slashes here. BC is the same as that length
is the same as that length. And so all four sides are
going to be that same length because this is a square. So let me write this down. I'll just write it this way. Because it's a square,
we know that CD is equal to BC,
which is equal to-- and we already
established this-- is equal to EC, which is equal
to EB, which is equal to EC. But the important thing
here is to realize that this and this
are the same length. And the reason why
that is interesting is it lets us know that this
is an isosceles triangle. So isosceles triangle, if
you have your two legs of it, the two base angles are
going to be congruent. So whatever angle
this green angle is, this angle is going
to be as well. So if somehow we can figure
out this angle right over here, we can subtract that from
180, and then divide by 2 to figure out these two. Because we know that
they're the same. So how can we figure
out this angle? Well, we know all the
angles of this thing. We can figure out the angles of
all of this larger one up here. We know this is an
equilateral triangle. So this over here has to
be 60 degrees, as well. That's 60 degrees. And that is also 60 degrees. In fact, I could
write over here, which is equal to the
measure of angle BCE. So if this is 60
degrees, we know we're dealing with a square,
so this whole angle over here is a right angle. What is the measure
of angle ECD? What is this angle
right over here? Let me do this in a new color. This right over here is going
to have to be 30 degrees, so that is going
to be 30 degrees. And now, we're ready to solve
for these two base angles. If we call these x-- and we
know they have to be the same-- we have x plus x
plus 30 degrees. x plus x plus 30
degrees is going to be equal to 180 degrees. That's the sum of all of the
interior angles of a triangle. So you get 2x plus 30
is equal to 180 degrees. Now, you can subtract
30 from both sides. And we are left with
2x is equal to 150. Divide both sides by 2,
you get x is equal to 75. So we've figured out
that x is equal to 75. And now we're at
the home stretch. We need to figure out angle BED. So x is equal to the
measure of angle CED. So BED is CED plus BEC, so the
60 degrees plus 75 degrees. So it's going to be-- ready for
the drum roll-- this is going to be equal to 75 degrees
plus 60 degrees, which is equal to 135 degrees. And we are done.