The diagram: it is a large triangle labelled ACD(AC a line horizontal, then AD and CD joining at the tip going down-so the triangle is "upside down" with the base up) with a line drawn that connects point E, which intersects segment CD at the middle(a median), with point B, which intersects line AC at the middle.

1. Question: In the figure above, line AD is parallel to line BE, and the area of quadrilateral ABED is three times the area of triangle BCE. What is the value of AD/BE(aka the two sides parallel)?
(I am inquiring about a general concept/formula/rule about a certain type of question through this question that I am stuck on- but I am unable to provide an image of the problem's diagram on this website so I hope someone can understand what I am trying to ask through my descriptions.
I think it is something about the ratio of sides,in this case 1/2(the square root of the area- a rule?) being the square root of the ratio of area?(which is 1/4 says my teacher, but why is it 4 and not 3 when one triangle's area is x and the other 3x?)The answer is two but I do not understand it.

This problem comes in many guises. One can be found here:

http://math.stackexchange.com/questions/105084/a-line-which-bisects-two-sides-of-a-triangle-is-parallel-to-the-third

Naturally, the area of the small triangle CBE is 1/4 the area of CAD, since it has half the base and half the altitude.

Wait, I have a logic problem. How does it equal 1/4 again? Could you please explain Steve?

the large triangle has base b and height h. So, its area is bh/2

The smaller triangle has base b/2 and height h/2, so its area is (b/2)(h/2)/2 = bh/8 = 1/4 (bh/2)

To find the value of AD/BE, we can use the concept of similar triangles.

First, let's label the intersection point of segment CD and line BE as F.

Since line AD is parallel to line BE, we can conclude that triangle BCF is similar to triangle ADF. This means that the corresponding sides of these triangles are proportional.

Let's denote the length of AD as x and the length of BE as y. Then, we have the following ratios:

BC/AD = CF/DF = 1/2 (as BD is a median and cuts CD in half)
BC/BE = CF/EF = 1/2 (as BE is also a median and cuts AC in half)

Now, let's use the area information given in the problem. The area of quadrilateral ABED is three times the area of triangle BCE. This can be represented as:

Area of ABED = 3 * Area of BCE

The area of a triangle is equal to half the base times the height. Hence,
(x + y)/2 * CF = 3 * (x/2 * BC)

Using the ratios we found earlier, we can substitute CF and BC in terms of AD and BE:

(x + y)/2 * (1/2 * AD) = 3 * (x/2 * (1/2 * BE))

Simplifying this equation, we get:

(x + y)/4 * AD = 3 * x/4 * BE

Dividing both sides by (x + y)/4, we have:

AD = 3 * x/3 = x
BE

Hence, AD/BE = x/y = 1/2.

Therefore, the value of AD/BE is 1/2.