Simple area and angles of squares and triangels

Question

This is a question APPARANTLY tested on primary 4 and I am in Sec 2,wondering how to do this question....None of my classmates also could finish the question.

Question:

ABCD and BFGE are squares.AE and BE meet at the intersection point where E is midway of DC.In square BFGE,it have 4 congruent triangles and a shape as shown in the figure below.Answer(Teacher challenged us:NO FIGURE SHOWN BELOW,DRAW YOURSELF)

(a)The ratio of ABCD to the smallest square

(b)Determine if the area of BCE equals to the area of the smallest square.

(c)Find the value of angle AEB.If possible,avoid using trigonometry.

First,the figure doesn't seem to be enough description to draw.Second,no measurements are given at all,which is literally impossible to use trigonometry.

I thought this question is a crap question given by teacher as impossible as a joke.I would love to see how would u have done it.

Here is what I have drawn for this:

Answer 1

If we take your figure, we can determine all required quantities.

1. The congruent triangles are right triangles, and their legs have a ratio of $1:2$. The former is because one such triangle is in the corner of the square, while the latter is due to the fact that $E$ is the midpoint of $CD$. Since all questions are invariant to scaling, you can choose your scale of length arbitrarily, e.g. in such a way that the short legs of the congruent triangles have length $1$. Then you'd have $AB=BC=CD=DA=2$ and $EC=CK=KL=LM=MC=1$ is half that length, so the small square has half the edge langth and a quarter the area of the square $ABCD$.

2. Well, the area of a right triangle is half the product of its leg lengths, in this case $1$ and $2$ as stated above. So yes, they are of equal area.

3. I don't know how to avoid trigonometry here. The triangle $ABE$ has one side of length $2$ and two sides of length $\sqrt{2^2+1^2}=\sqrt5$ according to Pythagoras' theorem. By the cosine law, you have $$\angle AEB=\arccos\frac{5+5-4}{2\cdot\sqrt5\cdot\sqrt5}=\arccos\frac35\approx53.13°$$ but that angle seems not to be a rational multiple of $\pi$ so I see no easy way to obtain that angle exactly without using trigonometric functions.