Similar triangles question If I have a right triangle with sides $a$.$b$, and $c$ with $a$ being the hypotenuse and another right triangle with sides $d$, $e$, and $f$ with $d$ being the hypotenuse and $d$ has a length $x$ times that of $a$ with $x \in \mathbb R$, is it necessary for $e$ and $f$ to have a length $x$ times that of $b$ and $c$ respectively?

EDIT: The corresponding non-right angles of both triangles are the same. 

As the comments say, no
Think about a triangle $ABC$ with hypotenuse $A$ and then another triangle $DEF$ with hypotenuse $D$. We know that there's a relationship between $A$ and $D$, so it's as if $D$ is given to us. You can imagine extending (or retracting) one of the legs on $DEF$ to some factor of the corresponding leg on $ABC$ and easily find another leg to complete the triangle.
Now let's say you wanted to keep the triangles as right triangles. Then it's clear that $A^2 = B^2 + C^2$ and that $D^2 = E^2 + F^2$. We know that $D = xA$ so we know $x^2A^2 = E^2 + F^2 = x^2 \left( B^2 + C^2 \right)$ so it follows that $\frac{E^2 + F^2}{x^2} = B^2 + C^2$. Now if we are initially given the $ABC$ triangle then we know $B , C$. It then follows that $\frac{E^2 + F^2}{x^2} = B^2 + C^2$ has many different solutions for a given $x$ (at first glance it appears to be an infinite number but I wouldn't quote me on that).
Hope this helps!
EDIT:
With the addition to the original question that all angles are the same then the answer is yes, all sides will be scaled equally. Let's call $B$ the side corresponding to $E$ and $C$ to $F$. Next let's call the angle between $A,B$ as $\theta$ (and similarly the angle between $D,E$). It must be true that $\cos \theta = \frac{B}{A} = \frac{E}{D}$. We take note that $D = xA$ so $\frac{B}{A} = \frac{E}{xA} \iff xB = E$. The same argument can be applied to side $C$ and $F$ using the angle between $A,C$ (and $D,F$).
Again I hope this helps!
