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## Geometry (FL B.E.S.T.)

### Course: Geometry (FL B.E.S.T.)>Unit 5

Lesson 3: Proofs of general theorems

# Geometry proof problem: squared circle

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?
• 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.
• 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?
• Nope. A triangle must have 3 straight lines with interior angles adding up to 180 degrees.
Happy to help!
• 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)?
• 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!
• At , 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?
• They are both 75 because it is an isosceles triangle (two of its sides and two of its interior angles are the same)
• How often will we use proof theorems/problems in real life?
• The math itself may not be too useful, but the critical thinking and logic is.
• At wouldn'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.
• Would the triangles at be using the postulate RSH?
• At we 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
• At I 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.