Formulate the opposite theorem to the first criterion for the equality of triangles and prove it.

The first sign of equality of triangles says that two triangles are equal when the two sides and the angle between them in these triangles are respectively equal.

Theorem:

If two sides and the angle between them of one triangle are respectively not equal to the two sides and the angle between them of another triangle, then such triangles are not equal.

Equal figures are those figures that, when moving, pass one into another. When moving, the angles between the rays and the lengths of the segments are preserved.

If △ ABC and △ A′B′C ′ are given, then during motion AB = A′B ′, BC = B′C ′, AC = A′C ′, ∠A = ∠A ′, ∠B = ∠B ′ , ∠C = ∠C ′.

Since the two sides and the angle between them of one triangle are not equal to the two sides and the angle between them of another triangle, then:

AB ≠ A′B ′;

AC ≠ A′C ′;

∠A ≠ ∠A ′.

Three out of six equalities are not true; hence, △ ABC ≠ △ A′B′C ′, as required.