Finding Ratio of Triangle Areas
The given problem is to find the ratio of areas of two triangles: △AFB and △BDE.
Let’s analyze the information given in the problem:
- △ABE is a right-angled triangle with ∠ACB = 90°.
- D and E are points on line segments BC and AC, respectively.
- BD : DC = 2 : 1, meaning BD is twice as long as DC.
- DC = EC = AD (since it's given that DC = EC and also DC = AD).
- BE and AD intersect at F.
From the given information, since BD : DC = 2 : 1, we can represent BD as 2x and DC as x for some value of x. This means BC = BD + DC = 2x + x = 3x.
Given that DC = EC = AD, we can also represent each of these segments as x.
Now let's find the ratio of areas of △AFB and △BDE. Note that these triangles share the same height if we consider the height to be extending from B perpendicular to AE. Therefore, the ratio of their areas will be proportional to the ratio of their bases AE and DE.
Since △AFB's base AE includes AD + DE and AD = x, we have base AE = x + DE. We now need to find the length of DE. Since E is a point on AC and EC = x, AE (which is the sum of AD and DE, and we already know that AD = x) must be 2x in length since AC is the entire hypotenuse which would be twice the smaller segment EC given △ACE is a 45-45-90 right triangle due to the equality EC = AD and those are the adjacent sides of the right angle.
Now we know that AE = 2x and DE is the remaining portion of AE which would equal x (since AE is 2x and AD is x, DE must be x to satisfy AE = AD + DE).
Therefore, the ratio of the areas of △AFB to △BDE is proportional to their bases, which is AF to FB or AE to DE. Since AE = 2x and DE = x, the ratio of the areas is:
Area(△AFB) : Area(△BDE) = AE : DE
Area(△AFB) : Area(△BDE) = 2x : x
Area(△AFB) : Area(△BDE) = 2 : 1
Thus, the ratio of △AFB to △BDE is 2:1.
