Do this geometries. It R Harder
-
Easiest way is to extend BD out to D' such that BCD' is isosceles, then you know the angles of regular trapezoid BEDD', which gives you x.
I can't follow what you mean without an actual diagram. When I try to do as you suggest, I find that D and D' are the same point.
-
Can we get extra credit for selfies with sex toys?
Only if our parents are equilateral.
-
Easiest way is to extend BD out to D' such that BCD' is isosceles, then you know the angles of regular trapezoid BEDD', which gives you x.
Not sure what you're getting at. If you're extending BD and placing D' anywhere on the extension, then BDD' is a straight angle, and BEDD' will never be a trapezoid.
-
I find that D and D' are the same point.
Indeed, BCD is already isosceles1; there is no need to introduce a D'. Not for that, anyway.1That much of my attempt was correct. I drew an embarrassingly incorrect conclusion from that, but I was correct up to that point.
-
@Buddy said:
Easiest way is to extend BD out to D' such that BCD' is isosceles, then you know the angles of regular trapezoid BEDD', which gives you x.
I can't follow what you mean without an actual diagram. When I try to do as you suggest, I find that D and D' are the same point.
@Buddy said:
Easiest way is to extend BD out to D' such that BCD' is isosceles, then you know the angles of regular trapezoid BEDD', which gives you x.
Not sure what you're getting at. If you're extending BD and placing D' anywhere on the extension, then BDD' is a straight angle, and BEDD' will never be a trapezoid.
Shit. CEDD'
@flabdablet said:
I find that D and D' are the same point.
Indeed, BCD is already isosceles1; there is no need to introduce a D'. Not for that, anyway.1That much of my attempt was correct. I drew an embarrassingly incorrect conclusion from that, but I was correct up to that point.
Were we looking at different diagrams? I think this one was posted by @accalia, so if the letters are all in the wrong place that's no surprise.
-
this one was posted by @accalia, so if the letters are all in the wrong place that's no surprise.
Were we looking at different diagrams?
No, same diagram. Angles DBC and DCB are both 20°; therefore, triangle DBC is isosceles.Shit. CEDD'
That I might buy. I haven't worked through your analysis to see if it's correct, but that is at least plausible.
-
Proof's a bit terse. I don't think you mentioned why angles CED and D'DE are equal?
Naturally BCD' and BD'C are equal to 80°, and if you can prove the trapezoid then BED is also 80°, and x would be 50°. Which mismatches @accalia's solution so I'm suspicious.
-
