Circle Theorems Puzzle

[Coontang]

We're doing circle theorems at school and one question I just couldn't work it out for half an hour. It bugged me so much I thought I'd post it here!

Picture quality isn't fantastic but you should hopefuly be able to understand it.

EDIT: By the way, the longest line is the diameter, and where it splits off (the three lines with the thin lines going across them) is the origin, so obviously the 3 lines i have marked are radius'.

[[SoE]_Zaitsev]

/Skips this topic

[Hoogie]

It's not clear to me which corner is the 64 degree one. Is it the corner the arrow is pointing or the 2 corners where the red is one of them?

[Drofder2004]

Use a protractor you fucking morons.

[Rezil]

It's 26. Posting proof in a sec.

[Drofder2004]

It's 26. Posting proof in a sec.

I think you will find you are wrong.

Because if you were right, the answer to the question which was posed is simply "Yes."

Can I get a medal for a correct answer?

[Rezil]

EDITED, also I love being a mod.

Anyway, proof:



1. Triangle ACE: Two equal sides, so the corners must be the same value. The top corner is 128, so its 180-128=52/2=26.

2. Triangle ABF: Corner BFA is 90 degrees(simmilar triangles: BCD and EFD - the rule is that simmilar triangles have the same corners). So 90+26=116, 180-116=64, and we already know that one of the smaller corners is 26, so 64-26=38.

3. The direct way is you seeing that ABF and ADF are simmilar(two equal corners and one side automatically means they're simmilar), corner ABF is 26 so corner ADF MUST be 26(the shit we're looking for).

[Drofder2004]

EDITED, also I love being a mod.

I have a screenshot of you being totaly wrong!

[Hoogie]

First it was 26 then 38 and now 26 again? xD

But 26 can't be right as the red angle is clearly bigger then the other half. So it has to be above 32.

[Rezil]

See above. Makes sense to me anyway.

[Pedsdude]

2. Triangle ABF: Corner BFA is 90 degrees(simmilar triangles: BCD and EFD - the rule is that simmilar triangles have the same corners).

I don't think you can say this. CBD and EFD are not similar triangles, because you don't know that AE (or maybe more clearly , FE) and BC are parallel.

EDIT: Oh nvm, forgot that it's 26+26=52 in the bottom right angle, so yeah, your proof seems correct

EDIT 2: No, wait, you only find out that the very bottom right angle is 26 AFTER assuming that BFA is 90 degrees, I believe.

[Rezil]

Shit, I know I overlooked something. Brb.

[Pedsdude]

I completely agree up to this point (see attachment).

[Pedsdude]

OK, looking at the image in my previous post...

Give the radius a length of 1. Then looking at the triangle ACD, you can work out the length of AD. You've got that the angle CDA is 90 degrees, which makes CD the hypotenuse, AD opposite, and CA adjacent. Therefore, using soh-cah-toa, sin(26) = opposite/hypotenuse = AD/2, so AD = 2sin26 = 0.87674...

(back later, got a CoD4 war, will continue my thought process later )

[Rezil]

Yeah but you don't have any lengths. So it's corners only.

Also, it IS 26, I just have to prove that ABF is 90°.