What we do

[tender laurel]

help

x=?

find

[earnest laurel]

is there more to this?

maybe a length

[tender laurel]

ABCD rectangle

AC and BD diagonal

just these

[earnest laurel]

you can find abe by 90-54 and then bec by 180-54+45
then aeb with that you can find bae

[tender laurel]

there is a golden ratio

[earnest laurel]

kinda stuck after that

what do the equal sides even mean

[high hollow]

idk wtf im doing trying to do this

[tender laurel]

whaat

[high hollow]

u can get a few angles but then like what

[earnest laurel]

i don't think I'm even near it

is this the right answer or 24

[tender laurel]

True answer

18

i want diffrent syntetic solutiob

[earnest laurel]

there's one with trigonometry

[tender laurel]

i dont wnna use trigo

[high hollow]

perhaps you could assign a length to one of the sides and use combinations of like angles and side lengths of a triangle to get other angles and lengths

like use stuff like side angle side and like all the other versions of that

[tender laurel]

i can try but i dont think so i cant solve 😭😭

[high hollow]

i mean tbh you should do it how the prof shows tho

if they're giving u a solution

not how random josh is guessing 💀

[tender laurel]

where is prof?

where is he

no prof

[high hollow]

i mean your teacher

[tender laurel]

haaa okaay

but i dont have a teacher

[high hollow]

💀 i would highly recommend taking math in person for like all math classes

[tender laurel]

@valid stag this guy very good

[tender laurel]

thanks i was

but school is alrd over

[valid stag]

This picture solves your problem if you prove that AF, HB and KC are concurrent.

[tender laurel]

oommmggggb

u r very gooood thanks

[tender laurel]

like this?

yess

[valid stag]

yes, but I don't see how one can quickly prove the concurrency of the yellow segments

[tender laurel]

i dont know either but my friend used the golden raito

.

[valid stag]

btw BK is not congruent to EK

while FE=FD

[tender laurel]

When u drww the radius it becomes decagon why do the sides have to be equal?

[valid stag]

because of symmetry

[tender laurel]

ohhh...okaayyy

[valid stag]

they are radiuses of the circumference

[tender laurel]

yes i understand this

[high hollow]

what in the world 🥀

[tender laurel]

@swift swallow

[swift swallow]

ok i see

did you alr figure it out

[tender laurel]

no proof

[swift swallow]

ok I have to leave

I have to go somewhere

[tender laurel]

ok good by

[swift swallow]

I’m back

In 5 min

Ok I’m back

ABK is 36 due to rectangle right angle then supplement theorem

ADB is 54 due to alt interior

Then BDC alt interior makes it 36

Then now we have a nice triangle

Man I can’t type this out i keep misspelling stuff I’ll do it on paper when I get home because I like this question it’s fun it’s like a proof that doesn’t end too quickly

[valid stag]

Oh, now I see how you can easily finish the proof from my diagram. Let O1 be the intersection point of HB and KC, then HO1/BO1=HK/BC.
Now let O2 be the intersection point of HB and AF, then HO2/BO2=HF/BO2=LF/AB because of similarity of triangles. So, O1 and O2 coincide.

[tender laurel]

https://tenor.com/view/h-o-l-y-s-h-i-t-boom-yes-oh-yes-yep-gif-15997893301937763341

Yesss correecctt!!