Problem about right triangles. Given N>1 right triangles. Sum one legs of each of them, then sum all the left legs, then sum all hypotenuses. These 3 sums form the sides of a right triangle. Prove all given N triangles are similar
I'm really stuck on this one. Got that sum of all hypotenuses is the end hypotenuse. but can't proceed from it. Any help would be appreciated.
 A: Consider the contrapositive statement:

If the right triangles aren't similar, then their edge-sums don't form a right triangle. 

Let triangle $i$ have legs $a_i$ and $b_i$, and hypotenuse $c_i$. We may assume that our triangles are ordered such that
$$\frac{b_i}{a_i} \;\leq\; \frac{b_{i+1}}{a_{i+1}}$$
That is, laying them out as in the figure, the hypotenuses have steeper and steeper "slopes" as $i$ increases:

The combined legs form legs of a large right triangle with hypotenuse $d$. The chain of hypotenuses form a polygonal path from one endpoint of the large hypotenuse to the other; clearly(?), then
$$d \;\leq\; c_1 + c_2 + c_3$$
with equality if and only if each $c_1$ lies on $d$. (We can make this more formal by dropping perpendiculars from the endpoints of the $c_i$ onto $d$, sub-dividing the large hypotenuse into segments of length $d_i$, where clearly(!) $d_i \leq c_i$ by the fact that a leg (here, $d_i$) of a right triangle is no longer than the hypotenuse ($c_i$).)
A: I can only give the following suggestion and its partial proof.

We will prove the claim by induction.
Let consider the case when $N = 2$. Also, we will restrict all angles to be less than or equal to $90^0$.
Then, according to the given hypothesis, we have $(a_1 + a_2)^2 + (b_1 + b_2)^2 = a^2 + b^2 = c^2$
After expansion and cancellation, we have $\frac {a_1a_2}{c_1c_2} + \frac {b_1b_2}{c_1c_2} = 1$
i. e. we have $\frac {a_1}{c_1}\frac {a_2}{c_2} + \frac {b_1}{c_1}\frac {b_2}{c_2} = 1$
i.e. we have $(\cos B_1)(\cos B_2) + (\sin B_1)(\sin B_2) = 1$
For the above to be true, B has to equal to B’ according to the identity ($\sin^2 x + \cos^2 x = 1$).
Thus, $⊿A_1B_1C_1$ ~ $⊿A_2B_2C_2$.
Now assume that it is true for $N = k$
We are now left with 3 triangles (because all others are similar already), namely  $⊿A_{k+1}B_{k+1}C_{k+1}$, $⊿A_{S(k)}B_{S(k)}C_{S(k)}$, and $⊿A’B’C’$ with the following explanation of the notation used.
The side opposing $A_{S(k)}$ has length = sum of all $a_i$ for $i = 1, 2, 3, .... k$. $B_{S(k)}$ and $C_{S(k)}$ should be interpreted similarly. 
The side opposing A' has length = sum of all $a_i$ for $i = 1, 2, 3, .... k, k+1$. Therefore, it length is equal to $S(k) + a_{k+1}$. B' and C' should be interpreted similarly.
The argument should be (and I hope) no different than the above in completing the induction proof.
A: The solution by Blue is very nice. Here's much the same proof, but with equations instead of the pretty picture. 
The triangle inequality says $$|v_1+\cdots+v_n|\le|v_1|+\cdots+|v_n|$$ with equality if and only if the $v_i$ are all scalar multiples of each other. Now let $v_i=(a_i,b_i)$, and let $a_i^2+b_i^2=c_i^2$, so $|v_i|=c_i$. Then $$v_1+\cdots+v_n=(a_1+\cdots+a_n,b_1+\cdots+b_n)$$ and the triangle inequality becomes $$\sqrt{(a_1+\cdots+a_n)^2+(b_1+\cdots+b_n)^2}\le c_1+\cdots+c_n$$ Squaring both sides, we have $$(a_1+\cdots+a_n)^2+(b_1+\cdots+b_n)^2\le(c_1+\cdots+c_n)^2$$ with equality if and only if all the $(a_i,b_i)$ are scalar multiples of each other, which is to say, if and only if all the $(a_i,b_i,c_i)$ right triangles are similar, and we're done. 
