Can there be two triangles, for example $ \triangle ABC $ and $ \triangle DEF$ so that the measure of their sides is the same but their angles are different? Thanks.
[Math] Is it possible to have two triangles with equal sides but with different angles
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The answer is no. See the following counter-example. The two triangles have the same altitude, and equal bases (and hence equal in area) but the third sides (i.e. BC, EF) are different.
This fact can also be verified by applying the formula:- area of a triangle = 0.5 ab sin C.
You can't prove it. In defining what is meant by have a consistent and measurable concept of "area" it is assumed that congruent shapes (having equal measurements and "able to fit ontop of each other") will have the same area. That is what "area" means.
So your proof is circular. Hurons formula (or the standard Area = half base times height) are all derived with the assumption that congruent shapes have the same area.
Consider the basic concept of Area = $\frac 12$ base $\times$ height. Why? Well, because if we have to congruent triangles we can form a parallelogram out of the them. The parallelogram has area base $\times$ height (why?). And so each triangle has half that area (Wait! Why do we assume the to congruent triangles have the same area?)
And you prove that a parallelogram has the area base times height because if you cut of a right triangle from one side to and move it to the other you have a rectangle that has sides that are length base and length height. So the area of the rectangle is base $\times$ height. (But 1; why does cutting the parallelogram and rearranging the pieces keep the area the same?; 2; why is the area of a rectangle base $\times $ height.)
Well, cutting a parallelogram and moving the pieces should keep the areas the same. (Why? Why shouldn't a triangle have one area when it is in one position and a different one in another?) And rectangle can be cut into a grid of squares. A square of $1 \times 1$ must have an area of $1$. (Why? Why should two congruent squares have the same area? And why does a $1 \times 1$ square have an area of $1$ unit?)
ALL of this is based on a single concept: That the quality that we call "area" is an aspect of dimensional lengths and angles. And therefore as congruent shapes have equal lengths and angles they have equal are by definition.
That's it. There is NOTHING to prove.
Best Answer
Yes, if the triangles aren't necessarily both Euclidean. This may not be what you meant, and the other answers may answer your question more appropriately, but this seems a good opportunity to point out some more exotic scenarios that can exist.
One thing we could do is take triangles from different geometries. For instance, the sum of the angles of a triangle drawn on a sphere will always be greater than $180$ degrees. So you could draw one triangle on a flat plane, another on a sphere, make the side lengths match, and you will get different sized angles.
Be aware though that it isn't sufficient to just scale the metric on a Euclidean space. That is, if I define two different planes where distance is greater in one but they are both flat, and I draw two triangles with matching side length, the angles will still be the same. The important thing about the example of the previous paragraph is that we changed the curvature of the space.
Maybe it's cheating to allow the two triangles to come from different geometries. Well it is in fact possible to draw two triangles with matching side lengths and different angles, in the same geometry. You can do this in the hyperbolic plane by allowing vertices on the boundary of the plane.
The boundary of the hyperbolic plane is like a circle that is infinitely far away from any point in the space. If you put two vertices on this boundary, then pick whatever point you want within the space for your third vertex, each side of the triangle will have infinite length. So the lengths of the sides would be unaffected by where you choose to put your third vertex. Also, the two angles at the vertices on the boundary would actually be $0$ degrees, and the third one could be any number up to and including $180$ degrees, depending on where you put it! This wouldn't work in Euclidean space. You can also talk about a circle boundary of $\mathbb{R}^2$ infinitely far away, but there, you can't draw a straight line connecting arbitrary boundary points and passing through the plane.
There's lots of Google-able info out there (and in Wikipedia too) about the hyperbolic plane, if this concept interests you and you want to know more!