This is going to be an degree angle right over. We have to make sure that we have the corresponding vertices map up together.
B was the vertex that we did not have any angle for.
These notes will introduce and discuss them one by one, but they, and their standard initials, are summarised here at the start to indicate the structure of the following discussion: The four standard congruence tests for triangles Two triangles are congruent if: SSS: the three sides of one triangle are respectively equal to the three sides of the other triangle, or SAS: two sides and the included angle of one triangle are respectively equal to two sides and the included angle of the other triangle, or AAS: two angles and one side of one triangle are respectively equal to two angles and the matching side of the other triangle, or RHS: the hypotenuse and one side of one right-angled triangle are respectively equal to the hypotenuse and one side of the other right-angled triangle.
Proof: This was proved by using SAS to make "copies" of the two triangles side by side so that together they form a kite, including a diagonal. So point A right over here, that's where we have the degree angle.
But I'm guessing for this problem, they'll just already give us the angle. A triangle with three sides that are each equal in length to those of another triangle, for example, are congruent.
Constructing a triangle with three given sides When all three sides of a triangle are given, however, there is no longer any freedom of movement, and only one such triangle can be constructed up to congruence.
So we do not prove it but use it to prove other criteria. This one looks interesting. AAS Angle-Angle-Side : If two pairs of angles of two triangles are equal in measurement, and a pair of corresponding non-included sides are equal in length, then the triangles are congruent.
If you try to do this little exercise where you map everything to each other, you wouldn't be able to do it right over here.