If you are given the angles (30deg, 60deg, 90deg) then it just is. AD doesn't have to halve BC for it to work
Someone can probably explain better but for most purposes it's just something you memorize Another common one is a 45-45-90 triangle having the ratio 1:1:sqrt(2)

Also
>X is the center point of the whole triangle
Nope you cannot be sure about that since the angle of ACB isn't given. For all we know it could be stretched far to the left and is actually like 1deg

Ok having thought a bit more you can probably think of it as

sin of a nonright angle in a right triangle is the ratio opposite/adjacent
sin(30deg) also happens to be exactly 0.5
You can work out the rest from there

>>1578998
*opposite/hypotenuse, sorry

>>1578997
OP here
Length of each side is the same, 2.
Every corner is 60 degrees

that's just how an equilateral triangle works. if you bisect an angle, you bisect the opposite side.

>>1579015
Is there a formal proof for this axiom?

>>1578994
It's pretty much this >>1578996
When you're given hard numbers, the diagram is just a visual aid. The numbers are always true and that's what matters. Nobody cares if the center line is 5mm to the right and at a tiny diagonal angle as long as there's the 90 degree box symbol and a length (or a derivable length, like if you were just given 2 and sqrt 3, the only correct solution is 1 because numbers > diagram)

>>1579135
I think >>1579015 assumed that it was an equilateral triangle to start with, but your image does not assume that. It just provides a triangle with a side 2, one angle of 60 degrees, a height of sqrt(3) and a partial angle of 30 degrees. You could hypothetically grab point C and drag it leftwards or bring it closer to D, and as long as it's moved on that line the other known numbers won't change. If you take C a mile to the left, it would most certainly not be an equilateral triangle.
As for your question:
Assuming ABC is an equilateral triangle, AB=AC. You drop a height from point A to BC, marking the intersection at D. Angles ADB and ADC are 90 degrees, since it's a height. Angles ACD and ABD are 60 degrees from it being an equilateral triangle. Triangles ACD and ABD are congruent for having two pairs of identical sides (AD shared between both, AB=AC), and having 2 (technically 3) identical angles (ADB=ADC and ABD=ACD). The implications of congruency: ABD and ACD are the same triangle, but mirrored. Angles BAD and CAD are identical, meaning that the height is an angle bisector - angles BAD=CAD=30 degrees. Sides BD=CD, meaning the height is also a median, splitting BC into two identical sections.
This can be done with all three vertices due to symmetry, hence in an equilateral triangle each height is also the median and angle bisector.

I'll address the question in the original image in the following reply.

File: laws of sines and cosines.png (34 KB, 700x259)

34 KB PNG

>>1579222
I'd like to correct my wording/translation mistake, the term is "altitude" rather than "height".
>>1578994
Let's answer the question about the ABD triangle with angles 30-60-90, with the side opposite the 90 degrees angle being of length 2. Let's ignore point C because it's outside.

Are you familiar with the law of SINES? It says that in any triangle, the ratio of each side divided by the sine of its opposite angle is the same. That means that AB/sin(ADB) = BD/sin(BAD) = AD/sin(ABD). If we apply numbers to these equations, we get: 2/sin(90°) = BD/sin(30°) = AD/sin(60°). sin(90°)=1, so the ratio is just 2. sin(30°)=1/2, so BD/0.5=2 => BD=1. That's exactly what you were looking for. You can do the same with sin(60°) to find AD being sqrt(3). To verify your conclusion, you can do Pythagoras on the three lengths of a right angle and find that indeed AB^2=BD^2+AD^2 (4=1+3).

Center point of the whole triangle. This extra question is tricky, because triangles have several different centers, so you have to be more precise. If I'm not mistaken, the intersection of the three medians, called centroid, splits the medians in a 2:1 ratio. If X is supposed to be the centroid, then XD is half the length of AX.
There are also the orthocenter, the intersection of the three altitudes which might actually be outside the triangle, and the incenter, the intersection of the three angle bisectors, which is the center of the trapped circle, but has no inherent ratio constraint.
In the case of an equilateral triangle, the three aforementioned centers align, and due to the quality of the centroid, X would still split AD into a ratio of 2:1.

I'll be around if you have more questions.

>>1579223
>>1579222
Thank you for this anon ( abit late)

>>1578994
The center of the triangle (centroid) absolutely does NOT half the line AD. In fact, you will find that there's a 2:1 ratio between the segments partitioned by the perpendicular bisectors and the center.

I recommend the book "Geometry by It's History" it'll walk you through everything you need to know and more. Work through that and have AI explain anything you don't understand.