In a right triangle, can $a+b=c?$ I understand that due to the Pythagorean Theorem, $a^2+b^2=c^2$, given that $a$ and $b$ are legs of a right triangle and $c$ is the hypotenuse of the same right triangle. However, most of the time, $a+b\neq c$. What I have been wondering is, is there any set of values for $a$, $b$, and $c$ that make the statement $a+b=c$ true?
 A: Your question has nothing to do with right triangles. If $c=a+b$ in any triangle, then the vertex $C$ must lie on the segment $AB$, and this is called a "degenerate" triangle.
A: 
(Addendum: As noted as well by others, this should highlight that your question has little to do with right triangles, as this is in fact a property of triangles in general)
A: Like Alex M. said, this doesn't have much to do with right triangles, but it's worth knowing the more general fact, and there's an easy visual!
Imagine you have two sides of a triangle, like an adjustable protractor. As you increase the angle between them, the other side is getting longer. But it won't equal their sum until you have your protractor fully opened up to 180°—and then it's not much of a triangle anymore. You also might wonder if there are any triangles where $a-b=c$ If you decrease the angle between your two sides, the other side is getting shorter. But it won't equal their difference until your protractor is totally closed to 0°—and again, then it's not much of a triangle anymore! 
So if we have a triangle and know the length of two of the sides, we know that the length of the other side is less than their sum, and greater than their difference. 
A: From the pythagorean theorem, $a^2+b^2=c^2$. 
Suppose $a+b=c$, then
$$
a^2+b^2=(c)^2\iff a^2+b^2=(a+b)^2
$$
Can you take it from here?
A: Substituting $c = a + b$ into $a^2+b^2=c^2$ gives us $a^2+b^2=(a+b)^2$. Multiplying that out gives us $a^2+b^2=a^2+2ab+b^2$ which means $2ab=0$
So to satisfy both $c = a + b$ and $a^2+b^2=c^2$ either $a=0$ or $b=0$. 
The question then becomes a bit more philosophical. We have a result that satisfies the Pythagorean equation but can we really consider it a right angle triangle? Is a "side" of length zero really a side at all?
A: There is another way to formally answer this question, namely using the cosine law.
For any triangle, it is true that $a^2+b^2-2ab\cos\gamma = c^2$ where $\gamma$ is the angle between the $a$ and $b$ sides. On the other hand, if $c=a+b$ then $c^2=(a+b)^2=a^2+2ab+b^2$, thus
$$a^2+2ab+b^2 = a^2+b^2-2ab\cos\gamma,$$
which is possible only if $\cos\gamma=-1$ or $\gamma=\pi$, that is, 180 degrees, or if either $a$ or $b$ is zero. Either way, this is a degenerate triangle, as mentioned in Alex M's answer.
A: One of the principles that this question illustrates is that in Euclidean geometry any path from point A to point B is not shorter than a straight line from A to B, i.e. a straight line segment A->B is the shortest path from A to B. Any deviation made through a point C that is not on the segment A->B makes a path longer and the only way that (A->C)+(C->B) can be equal to (A->B) is that C is on that segment.
A: Let $x$ be the angle opposite to altitude.
Now,we want $a+b=c$ or
Dividing both sides by c we get---                   $a/c+b/c=1$
By definition $a/c=sinx$ and $b/c=cosx$
So equation becomes----
$sinx+cosx=1$ or $sinx+(1-(sinx)^2)=1$
By solving we get----
$2(sinx)^2-2sinx=0$
We get 2 solutions that is
$sinx=1$ and $sinx=0$
Or as principle solution we get
angle $x=0$ and $x=π/2$.
Which means $a/c=0$ or $a=0$,$b=c$ and
$a/c=1$ or $a=1$,$b=0$.
We conclude that either base or the altitude should tend to $0$.
A: No, it can't. Although in case of simple triangle, make any one angle 180 degree and then the area of triangle will be zero. It will look like straight line. 
A: It has nothing to do with right triangles. There are two line segments and a third one straddling them. Sometimes it is called a $ diangle,$ a needle like shape enclosing no area between them.
A: Maybe, the equations $a^2+b^2=c^2$ and $a+b=c$ can be hold simultaneously in a ring or finite field. Say, over $F_{2^m}$, if we set $c=a+b$, then $c^2=a^2+b^2$ always holds. In fact, we know that $(a+b)^p\equiv a^p+b^p\pmod p$ for any prime $p$.
