# 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.

• In a word, no. You never specified the non-right angles. – Ron Gordon May 24 '13 at 17:54
• Try keeping lengths $a$ and $d$ equal and constant (meaning you assume that $x=1$), can lengths $b$ and $e$ be changed in such a way that $b$ is not equal to $e$? – Jerry May 24 '13 at 17:54
• No. Counterexample with $a=d=5$, $x=1$, $b=c=\frac{5}{\sqrt{2}}$, $e=3$, $f=4$. – Angela Richardson May 24 '13 at 17:54
• A way of picturing the situation, which relates to Jerry's comment: if you had two ladders, one being $\ x \$ times longer than the other, and leaned them individually against the wall in any way they could remain stationary, are you guaranteed that the height of the points where they meet the wall, or the distance from the wall where their feet touch the floor, are also different by exactly a factor of $\ x \$ ? – colormegone May 24 '13 at 18:02
• Well, with the edition of the original question, all the answers are incorrect. Two triangles are similar iff their corresponding angles are the same. Hence two right triangles with the same corresponding non-right angles are similar. – Philippe Malot May 24 '13 at 18:10

## 3 Answers

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!

EDIT: This answer is now incorrect since the OP changed his question.

This is a good way of visualizing the failure of your claim:

Imagine the point C moving along the circle from A to the north pole. This gives you a continuum of non-similar right triangles with a given hypotenuse (in this case the diameter).

Triangles are similar if and only if their angles match up. For the triangles to be similar then, the right angle of the one triangle must match up with the right angle of the other (why?). This means that side $d$ pairs with side $a$.

But there are still two possibilities: $b$ and $c$ can pair up with $e$ and $f$, respectively, or they can pair up with $f$ and $e$, respectively. Your conjecture assumes $e = xb$ and $f = xc$, which is the case when $b, c$ pair up with $e, f$; but what about the other case when $b,c$ pair up with $f,e$?