I would just like to say

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My baby's "hand-drawn" pyrography on 3 oak chopping boards.
1725995267822.png

and...
1725995315469.png


and...
1725995376577.png


Isn't she a clever girl? :D
No stencils or "lasers" @My Old Landy ;)
 
Stanleysteamer said:
This one is for @kevstar primarily, seeing as he has done it recently. But I laughed until the tears ran down my face, then asked the question! At the bottom.

"Just to put a toe in the water so to speak , In November I will be replacing my kitchen , therefore I wonder as it is in good condition ,made by a company called Elizabeth Anne , shaker styleIncluding a matching Neff dishwasher. Is any one interested in having it for about £1 ,500 Ono? Photos will appear , viewing always available.


cecilia d.
•Purewell & Stanpit•1d

A bargain for someone if they can fit it themselves!
Like

John H.
•Ameysford•Now

Is that the price removed or would you like us to take it all out for you?
Like
🤣 🤣 🤣 🤣 🤣 🤣 🤣
Click to expand...

I hate kitchens 🤣 🤣
 
Stanleysteamer said:
So I woz right first time then? Flipping miracle.
I wasn't sure that A-D and D-E would be equal as although the angle was bisected, the triangle is not isoceles, so if the sides are of different lengths then the two bits of A-E might not be the same, hence my remark.

Eventually I'll cheat and look it all up on the site the puzzle comes from. ;)
Click to expand...

I will just whizz past em cause ive got other things id rather do then fry my matter, id rather have somit in batter. 🤣🤣
 
DanClarke said:
OK, No trig needed possible because no lengths are stated.
Pure geometry problem.
D is 80° therefore the other side of D (BDA) is 100°.
Angle BAE is 70°
Angle BCA must be 50°
A-D and D-E must be equal (each is rooted on a 10° angle to a vertical line).
So angle ECD and ACD must be equal. So X is 25° (half of angle BCA).

Best answer I have.
Click to expand...
Even those who comment on the problem, and usually do get it right, most of them, cannot agree.


I was always taught in Maths to exaggerate a case to see where it leads.
So, imagine if AE was much much longer. With the angles at B, even though always equal, would their bisector always touch AE at its mid point? I somehow don't think so. I think ED would always be shorter than DA, As I said, BAE is not an isoceles triangle. Only if AB=AE could we be sure that the angle at A bisector would always divide AE equally.
 
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