Why would two triangles with exact same lengths have the exact same 3 angles? I know how incredibly stupid this question sounds, and I know any 2 triangles you draw with same line lengths will have the same 3 angles, but I just can't figure out the exact reason why having 3 lines of the same, or congruent, lengths, would lead to 2 triangles having the same angles. Can someone please explain?
 A: 
No matter what, in order for $a$ to be the same length, point $C$ must fall on circle $d$.  And no matter what, in order for $b$ to be the same length, point $C$ must fall on circle $e$.
With these restrictions, the only choices for $C$ are the two points where the circles intersect.  And these choices differ only by being mirror images of each other, which doesn't change angles, only their winding.
A: If you are comfortable accepting the law of cosines, that provides one explanation.  This theorem states that, if a triangle has sides with lengths $a, b$, and $c$ whose opposing angle measures are $\alpha, \beta$, and $\gamma$ respectively, then the following relation holds:
$$c^2 = a^2 + b^2 - 2ab \cos(\gamma)$$
Of course, we can arrive at two similar equations since there are $3$ different ways to label the sides with those letters and whichever way we choose is arbitrary.  For instance, we also have:
$$b^2 = a^2 + c^2 - 2ac \cos(\beta)$$
Looking at the first, notice that we can solve for $\cos(\gamma)$ in terms of the side lengths.  Because $\cos:(0,\pi) \rightarrow [-1,1]$ is an injective function, the value of $\cos(\gamma)$ determines $\gamma$ itself.  To be explicit:
$$\gamma = \arccos \left( \frac{c^2 - a^2 - b^2}{-2ab} \right)$$
And we have similar equations for angles $\alpha$ and $\beta$.
A: Given three lengths $a,b,c$ such that $a+b\ge c$, $a+c\ge b$ and $b+c\ge a$ we can construct the triangle uniquely up to rotation, translation and reflection: Draw points $A,B$ at distance $c$ connected by a line segment. The third point $C$ must have distance $a$ from $B$ and distance $b$ from $A$. Hence, it is an intersection point of the circle around $A$ with radius $b$ and the circle around $B$ with radius $a$. There are exactly two such points and the resulting triangles are mirror images of each other. The only choices here were where to put the line segment $AB$ and which of the two intersection points you choose. None of these choices influence the angles of the triangles.
