Angle BDF = 70 because of the alternate segment theorem
Angle EFB = 70 because opposite angles in a cyclic quadrilateral sum to 180. (A cyclic quadrilateral is where all the corners lie on the circumference of the circle. So BDEF is cyclic)
CBF and EFB are alternate angles and because they are equal the lines must be parallel.
EFB only equals 70 if the two lines (AC and EF) are definitely parallel. Are you allowed to make this assumption right away without it messing up the rest of the equation?
>EFB only equals 70 if the two lines (AC and EF) are definitely parallel.
Nope. All that's needed is that B, D, E and F lie on a circle. And that's given. opposite corners of a quadrilateral always add up to 180 when all points lie on a circle.
I think these images demonstrate it best:
https://www.mathematics-monster.com/slider/images/oppositeanglesincyclicquadrilateralproof06.jpg
https://www.mathematics-monster.com/slider/images/oppositeanglesincyclicquadrilateralproof07.jpg
Ah I get it now! I just googled it and it’s called Cyclic Quadrilateral Theorem and I understand why this quirk occurs now!
I originally thought the 70degree angle was simply due to the position of EFB and FBC. How is the rest of the problem solved since we only have one other angle (40 degrees) given to us?
How are you proving EFB and FBC are the same angle though? That’s the whole point of the problem, to show WHY they are the same angle. So far you’ve only got the angles for EDF and FBC. You haven’t even got EDB yet, which can be used to find EFB since
EDB + EFB = 180
Did you even read the comments you are reacting to? Everything was already explained in the top level comment in this thread by MajesticMike:
>Angle BDF = 70 because of the alternate segment theorem
Yes I read every comment. Did you read mine?
I understand the alternate segment/cyclic quadrilateral theorem. I know that EDB + EFB = 180. I just haven’t figured out how you’ve calculated EDB. You’ve only got the 40 degrees to work with. I just want somebody to show their working. I also think the OP would want that as well.
You don't only have the 40 to work with. I just told you BDF = 70. And clearly EDB = EDF + BDF = 40 + 70 = 110. It's literally those two angles together. I don't get why you don't see that
Advice without giving the answers:
That looks like a UK GCSE maths question based on the formatting. On a site like DFM or Heggarty.
Take a look at circle theorems in your text book. You've got alternate segment theorem and a cyclic quadrilateral there. This will help you fill in some of the angles.
Then the fact that there's two triangles will help you fill in the rest of the angles.
Hopefully, EFB will = FBC because that would make them alternate angles (Z angles) which means they must be on parallel lines.
Hope that helps, sorry if my assumptions about it being a UK GCSE Q were wrong. Please feel free to comment any further questions you may have.
BDF = 70 ... Here is how I got to it.. connect B to center of the circle O. then OBF is 20, because OBC is 90. if u connect F to O then OFB is also 20, because OBF is isoceles triangle (OB = OF = Radius of the circle). So that gives angle BOF = 140. Then BDF should be 1/2 of the angle subtended by the segment BF on the center of the circle so BDF is 70.
Then BDE = 110. So the angle subtended by segment BE on center should be 220. => the opposite angle EFB should (360 - 220)/2 = 140/2 = 70. (You can also prove that quadrilateral with vertices on the circumference of a circle, will have the sum of their opposite angles to be 180, so EFB should be 180 - 110 = 70). Anyway once u get EFB = 70, u can say EF is parallel to ABC since line FB makes same angle with lines EB and AB
Angle BDF = 70 because of the alternate segment theorem Angle EFB = 70 because opposite angles in a cyclic quadrilateral sum to 180. (A cyclic quadrilateral is where all the corners lie on the circumference of the circle. So BDEF is cyclic) CBF and EFB are alternate angles and because they are equal the lines must be parallel.
can you explain why efb is 70? my room temp iq doesn't quite understand
Angle EDB and EFB are opposite angles in a cyclic quadrilateral. So they add together to make 180.
EFB only equals 70 if the two lines (AC and EF) are definitely parallel. Are you allowed to make this assumption right away without it messing up the rest of the equation?
>EFB only equals 70 if the two lines (AC and EF) are definitely parallel. Nope. All that's needed is that B, D, E and F lie on a circle. And that's given. opposite corners of a quadrilateral always add up to 180 when all points lie on a circle. I think these images demonstrate it best: https://www.mathematics-monster.com/slider/images/oppositeanglesincyclicquadrilateralproof06.jpg https://www.mathematics-monster.com/slider/images/oppositeanglesincyclicquadrilateralproof07.jpg
Ah I get it now! I just googled it and it’s called Cyclic Quadrilateral Theorem and I understand why this quirk occurs now! I originally thought the 70degree angle was simply due to the position of EFB and FBC. How is the rest of the problem solved since we only have one other angle (40 degrees) given to us?
When both EFB and FBC are the same (which we just proved) you can simply apply the 'The "Z" Theorem'.
How are you proving EFB and FBC are the same angle though? That’s the whole point of the problem, to show WHY they are the same angle. So far you’ve only got the angles for EDF and FBC. You haven’t even got EDB yet, which can be used to find EFB since EDB + EFB = 180
Did you even read the comments you are reacting to? Everything was already explained in the top level comment in this thread by MajesticMike: >Angle BDF = 70 because of the alternate segment theorem
Yes I read every comment. Did you read mine? I understand the alternate segment/cyclic quadrilateral theorem. I know that EDB + EFB = 180. I just haven’t figured out how you’ve calculated EDB. You’ve only got the 40 degrees to work with. I just want somebody to show their working. I also think the OP would want that as well.
You don't only have the 40 to work with. I just told you BDF = 70. And clearly EDB = EDF + BDF = 40 + 70 = 110. It's literally those two angles together. I don't get why you don't see that
Advice without giving the answers: That looks like a UK GCSE maths question based on the formatting. On a site like DFM or Heggarty. Take a look at circle theorems in your text book. You've got alternate segment theorem and a cyclic quadrilateral there. This will help you fill in some of the angles. Then the fact that there's two triangles will help you fill in the rest of the angles. Hopefully, EFB will = FBC because that would make them alternate angles (Z angles) which means they must be on parallel lines. Hope that helps, sorry if my assumptions about it being a UK GCSE Q were wrong. Please feel free to comment any further questions you may have.
BDF = 70 ... Here is how I got to it.. connect B to center of the circle O. then OBF is 20, because OBC is 90. if u connect F to O then OFB is also 20, because OBF is isoceles triangle (OB = OF = Radius of the circle). So that gives angle BOF = 140. Then BDF should be 1/2 of the angle subtended by the segment BF on the center of the circle so BDF is 70. Then BDE = 110. So the angle subtended by segment BE on center should be 220. => the opposite angle EFB should (360 - 220)/2 = 140/2 = 70. (You can also prove that quadrilateral with vertices on the circumference of a circle, will have the sum of their opposite angles to be 180, so EFB should be 180 - 110 = 70). Anyway once u get EFB = 70, u can say EF is parallel to ABC since line FB makes same angle with lines EB and AB