Drawing squares on the sides of a right triangle,joining the vertices and so on... 
As shown at the figure: ABC is a right triangle.Firstly, we erect squares on its sides.Then we join the consequtive vertices of the squares and we get 3 more triangles.We continue and we erect 3 squares on the sides of these 3 triangles,we join the vertices and we get trapezoids and so on and so forth(next step we erect squares on the sides of the trapezoids)
I have managed to show that the sum of the areas of the three initial triangles is 3 times the area of the right triangle.Then the sum of the trapezoids is 5 times the area of the right triangle...So my question is: is there a pattern behind this procedure?Possibly if we go on we will always be getting multiples of the area of the right triangle.
 A: Let me call $s$ the area of the original triangle and $a$, $b$, $c$ the areas of the "first generation" triangles. We have $s=a=b=c$, because any triangle among $a$, $b$, $c$ has two sides of the same length as two sides of $s$, and the angle between them supplementary of the corresponding angle in $s$.

You can then for instance rotate $s$ 90° counterclockwise around the common vertex with $a$, to create a larger triangle with double area. But triangle $a$ has a vertex in common with a "second generation" quadrilateral (green in the diagram below): we can rotate that larger triangle 90° counterclockwise around the common vertex to create another quadrilateral. Finally, we can rotate $b$ 90° clockwise around the common vertex with the quadrilateral, to get another larger quadrilateral. This quadrilateral is a parallelogram, because its opposite short sides have the same length as the congruent sides of $s$ and $b$, while the angles adjacent to these sides are the same as the corresponding angles in $s$ and $b$ and thus supplementary.

This parallelogram has a the upper base which is four times the base of $s$, while its height is the same as that of $s$. Its area is then $8s$ and if we subtract from that the area $3s$ of the additional triangles we find that the area of a second generation quadrilateral (which is then a trapezoid) is $5s$.
A similar construction can be made for a third generation trapezoid. I won't write all the details but you can see from the diagram below that such a trapezoid (orange) can be completed into a parallelogram by adding two second generation trapezoids and six triangles. Such parallelogram has a base which is five times the base of $a$, and a height which is four times that of $a$. Its area is then $40s$ and subtracting from that the area $16s$ of the added figures we get $24s$ as the area of a third generation trapezoid.

These constructions can be carried out for every $n$-th generation trapezoid, but it is not clear how to derive from these a recurrence relation between their areas $S_n$. Apparently, though, such a recurrence can be easily guessed to be
$$
S_n=5S_{n-1}-S_{n-2}.
$$
EDIT.
It is indeed possible to obtain a recursive relation, by examining three trapezoids of successive generations (see diagram below). Let $S_n$ be the area of a $n$-th generation trapezoid, $a_n$ its smaller base, $b_n$ its longer base and $r_n=b_n/a_n$. The $n$-th trapezoid can be translated on top of the $n+2$-th to create a larger trapezoid, because $a_{n+2}=b_n$. By triangle similitude one has then 
$(b_{n+2}-a_n):(b_n-a_n)=(b_{n+1}+a_{n+1}):a_{n+1}$, whence:
$$
r_{n+2}=\left(1-{1\over r_n}\right)(1+r_{n+1})+{1\over r_n},
$$
a formula which allows to find recursively the values of $r_n$, given that $r_2=4$ and $r_3=5$:
$$
r_4=\frac{19}{4},\ r_5=\frac{24}{5},\ r_6=\frac{91}{19}, 
r_7=\frac{115}{24},\ r_8=\frac{436}{91},\ r_9=\frac{551}{115},\ 
r_{10}=\frac{2089}{436}
$$

In a similar way one can find a relation for $S_{n+2}$ in terms os $S_n$:
$$
S_{n+2}={r_{n+1}\over1+r_n}(r_n r_{n+1}+2r_n-r_{n+1})S_n,
$$
a formula which works for $n\ge2$. Starting with $S_2=5$ and $S_3=24$ one then finds:
$$
S_4=115,\ S_5=551,\ S_6=2640,\ S_7=12649,\ S_8=60605,\ 
S_9=290376,\ S_{10}=1391275.
$$
