Segar Rogers
@segarrogers.bsky.social
380 followers 340 following 530 posts
Teacher. Maths. Secondary. Edinburgh. Old enough to remember chalk. Poetry on a Sunday.
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segarrogers.bsky.social
Gotta love James Taunton … an amazing method – thanks :-)
segarrogers.bsky.social
I did not know this! Brilliant. Thank you :-)
segarrogers.bsky.social
I agree with you yes; there’s usually a ‘proof by demonstration’ … or a ‘proof by what’s the pattern’. I was interested in thinking about the things we gloss over because the pupils don’t have the mathematics to properly prove it yet. And of course it’s natural to climb on the shoulders of others.
segarrogers.bsky.social
True. I was just meaning on a very general level .. like using the formula for the volume of a sphere with say an S3 class … which would then need S6 Advanced Higher calculus to prove it.
segarrogers.bsky.social
lol … I don’t recall a ‘by symmetry’ argument in the Elements ;-)
segarrogers.bsky.social
As an aside, have you ever taught integration first? ( … as accumulation … area under a curve … it’s lovely!). You can then just about teach differentiation as anti-integration … er, admittedly by avoiding proving the fundamental theorem of calculus ;-)
segarrogers.bsky.social
Fair point … I think Gottlob Frege decided that we’d made them up … but don’t quote me on that ;-)
segarrogers.bsky.social
What things in maths do we teach before we have taught the maths that shows it is true? #UKMathsChat #iTeachMaths
segarrogers.bsky.social
Maybe repeat but in clockwise direction. Hmm? Thinking ;-)
segarrogers.bsky.social
So …

AB rotates through 50°.
∴ arcBB’ is 50°
∴ arcAA’ is 50°
∴ angleAOA’ = 50°

… then repeat (?) … hmm, not sure how to show A’ lands at C without invoking Eu.III–20?
segarrogers.bsky.social
This one is a little harder to see because the vertex of the 55° angle coincides with the end of the 110° arc. But the 110° arc is still being subtended from a point on the circumference, so Eu III-20 still applies.

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segarrogers.bsky.social
... and the same again, this time starting at the 40°

1 → 2: Eu. III, 20
2 → 3: Supplementary arc.
3 → 4: Eu. III, 20

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segarrogers.bsky.social
Start at the 20° and follow the sequence of steps.

1 → 2: Eu. III-20
2 → 3: Supplementary arc.
3 → 4: Eu. III-20

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segarrogers.bsky.social
Euclid's Elements Book 3 Proposition 20 (called by some the 'Inscribed Angle Theorem') will look like this: the arc is double the angle (on the circumference) that subtends it.

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segarrogers.bsky.social
An arc of 30° will have a supplementary arc of 150°.

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segarrogers.bsky.social
A very short primer on mixing degrees of arc with degrees of angle for circle theorems.
#UKMathsChat #iTeachMath

Historically degrees measured arcs and arcs measured angles. This might feel uncomfortable but remember how we measure an angle today … with a protractor … measuring around an arc.

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segarrogers.bsky.social
Ah, a jumble of half-remembered procedures – there’s a surprise!

The last thing that tends to occur to pupils is to first draw the situation.
segarrogers.bsky.social
Is that what you call them:’Angle-Arc Theorems’?
segarrogers.bsky.social
Lol … when I made the question I did think ‘Karen will know this … she wrote that post with all the formulas!!’.

What I worked out this time around was how to do it without the formula I.e. using only Eu.III, 21.
segarrogers.bsky.social
Yes, exactly. But try to avoid converting it back to what you know and work on the circumference rather than at the centre; all you really need to know is Euclid III, 21.
segarrogers.bsky.social
They’re the arc lengths (measured in degrees). Degrees measure arcs and arcs measure angles. I mix my arcs and angles (for convenience … it keeps the diagram less cluttered). I first saw it done a few years ago. Historically it’s correct. I can never go back to angles-only, I’d miss the flexibility.
segarrogers.bsky.social
That’s nicer ;-) Interesting language – ‘What fraction of the circumference’ … I was expecting degrees or degrees of arc.
segarrogers.bsky.social
Sorry, I forgot I wasn't talking to the first years ;-) Lets go for n, s ∈ ℕ/{0}