Khan Academy Needs Stronger Community Forum

I have come to find there is a 'Community Questions' link that is somewhat what I was looking for. However, this is quite hard to locate. 

When one goes tohttps://www.khanacademy.org/ what comes up? All the subjects one is involved in. When you click on the subject is there a link to 'Community Questions'? No. To find 'Community Questions' one needs to then click on 'See All________' then scroll to the very bottom of the page and there finally one has found a link to 'Community Questions' berried away. 

I think it would be a good idea to make 'Community Questions' a bit more obvious and visible.

Hi Neeziv - thanks so much for your interest in this Support Community!  I agree that this community could use some more love from a feature perspective.  Single sign-on between the main Khan Academy site and this forum is a planned feature that I hope we get to soon.  Most people here agree that having users sign up for an additional site is unnecessary and confusing, and that we should change it.  I've been working on getting the engineering resources prioritized for it, but that's never a straightforward process unfortunately (here or any tech company :-/).

As for questions and answers specifically about the content covered on the site, as you mentioned you can actually find a lot of discussion under the content pages themselves (articles and videos).  People are typically very helpful when someone has a question on the subject being covered.

In addition to supporting seamless sign-on between KA and this Support Community, there are various other projects around this part of the site that I hope to see get started over the coming weeks/months.  The KA community is one of the most helpful and friendly on the web, and we would love to be able to further facilitate people helping each other through their learning progression.

Anyways, thanks a ton for the feedback, I hope you keep your suggestions coming.

Comment : Khan Academy Needs Stronger Community Forum

Discussion:  Yes, I agree. I also the clear focus of the videos produced by Khan Academy people. But, learning is not just an individual task. It's social interaction task.  Community via internet is difficult due to impersonal nature of the people who participate. At urban hubs, Khan Academy could partner with a local hospital sponsor perhaps, to have meetup events where people could come and participate in a real community, say once a month.  Need example of a meetup.com "eyeball to eyeball" social dynamics oriented learning to help build community wealth ?  Example:  Community wealth building circle at slideshare.net  https://www.slideshare.net/icptwin45/community-wealth-building-circle-at-slidesharenet

Geometry is the most challenging subject and diverse subject on earth. I believe if taught properly is the best mechanism to prepare students for an engineering, science, or math major. Geometry overlaps with chemistry, physics, and engineering. Geometry is art with numbers, its creative, with the logical exercises below, you have to know how to draw and set up the proof, prove and given. Moreover, know the theorems. I am writing this material because most high schools in the USA do not teach this in an honors geometry course, I v spoken with several teachers, and some teachers use my material, but some on the other hand do not or cant teach this because, they don’t know how to teach the material, or its not in the state standards.

Prove a radius drawn to the midpoint of an arc bisects the chord of the arc and is perpendicular to it.

Prove A diameter with center O, to the midpoint of an arc bisects the chord of the arc and is perpendicular to it.

Prove: A line through any point within an angle, perpendicular to the bisector of the angle, makes equal angles with the sides.

If the perpendiculars drawn from the points of trisection of aside of a triangle to the other two sides are equal, the triangle is isosceles.

Prove any two altitudes of an equilateral triangle are equal.

Prove that two right triangles are congruent if the hypotenuses (side opposite the right angle) and side of one are equal respectively to the hypotenuse and side of the other.