Level 6 - Solution not accepted

[AsWaM]

I suspect it is the same problem as in my previous issue (level 5) as it relies on the same principle.

Step 1: Draw a circle centered on A, going through C, it crosses AB on point D Step 2: Draw a circle centered on D, going through C (and of course through A) , it crosses AB on point E Step 3: Draw a circle centered on E, going through D , it crosses the circle from step 2 on point E --> ACD and DEF are both equilateral, and have the same size, so the bissectrice of angles DFE and DAC are both perpendicular to AB. Moreover the distance between C and AB is the same as the distance between F and AB, so segments EF and AB are parallels.

[AsWaM]

I know the double perpendicular solution is more efficient, but I used only the circle for this one

[AbigailBuccaneer]

Step 2: Draw a circle centered on D, going through C (and of course through A) , it crosses AB on point E

Not a valid solution - the circle centered on D doesn't go through A - that would only be the case if the angle CAB were 60 degrees, which it is nearly , but not. You can just about see that in your screenshot.

Regardless, level 6 has been updated so it's not longer so close to 60 degrees, and you can see much more easily that this doesn't work.

[AsWaM]

Hi, I reckon the mistake, the ACD and DEF triangles are not equilateral, but only isoscele.

Thanks for your time, and sorry for the unnecessary trouble.

Alex