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# Geometry proof problem: squared circle

CCSS.Math:

## Video transcript

so we're told that quadrilateral ABCD is the square which tells us that the four sides have equal lengths and that they are all the interior angles are 90 degrees we also know that FG 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 this is the center of the circle this is part of 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 B edie is so what is B e D so it's B II D 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 I you could pause it and try it without any hints and now I will give you a hint if you if you tried it the first time and you weren't able to do it and you should pause again after this hint is 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 is to realize that this is a circle and so any any line that goes between B and any point on this arc is going to be equal to the radius of the circle so a B is equal to the radius of circle B E is equal to the radius of the circle and we can keep drawing other things that are equal to the radius of a circle B C is equal to the radius of the circle so let's think about a little bit if we were to draw in 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 hot thinking about how to do this problem so let me draw a segment EC and draw that as straight as possible I can draw a better job of that so segment EC now something becomes interesting because what is what is the relationship between triangle EBG and triangle ECG well they both definitely share they both share this side right over here they both share side EG and then B G is equal to G C 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 side angle side that these two triangles are going to be congruent so we know that triangle triangle e b G is congruent to triangle ECG II CG I should emphasize the C not the e ECG buy side-angle-side congruence by side-angle-side congruence and that also tells us that all of the corresponding angles and sides are going to be the same so that tells us that tells us right there that EC that EC is equal to EB EC is equal to EB so we know that EB EB is equal to EC and what is what also is equal to that length well once again this is the radius of the circle ve is 1 radius of the circle going from the centre to the arc but so is BC it is also radius of the circle going from the centre to the arc so this is also equal to this is also equal to BC so I could draw this other other three things right here I'm referring to the entire thing not just one of the segments all of BC so what is what kind of a triangle is this right over here triangle B EC triangle B e C is equilateral 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 B EC VEC we're not done yet but it gets us close is 60 degrees so the measure of angle B EC right over there sixty degrees so that gives us part of the problem B EC is part of the angle B e D 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 sixty degrees and we're done we figured out the entire the entire be e d 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 all four sides are going to be that same length because this is a square so let me write it let me write this down because because it's a square I'll just write it this way because it's a square we know that we know that CD is equal to BC which is equal to and we already establishes is equal to C 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 it's an isosceles triangle so whatever 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 somehow somehow we can figure out this angle right over here we can subtract that from 180 and then divide by two 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 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 measure of angle b.c.e BCE so this is 60 degrees we know we're dealing with the square so this whole angle over here so 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 if you know now we're ready to solve for these two base angles if we call these X if you 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 2x plus 30 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 figured out that X is equal to 75 and now we're at the home stretch we need to figure out angle B edie well that's going to be angle C so X is equal to the measure of angle CED so ve D is CED plus CED plus B EC 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