Final answer:
Triangle AEB in rhombus ABCD can be proven to be a right triangle due to the properties of a rhombus, where its diagonals are perpendicular bisectors and thus create right angles where they intersect.
Step-by-step explanation:
To prove that △AEB is a right triangle in a rhombus ABCD, we must first understand that the diagonals of a rhombus are perpendicular bisectors of each other. Therefore, if we look at diagonal BD, it would be cut into two equal segments at point E, which is the intersection of the diagonals. Additionally, diagonal AC would do the same, effectively splitting the rhombus into four congruent right triangles.
Specifically, in △AEB, AE and BE are half the length of the diagonals of the rhombus. We know that in a rhombus, all sides are of equal length (AB = BC = CD = DA). Taking into account the properties of the diagonals, which are perpendicular bisectors, we see that angle AEB must be a right angle. Therefore, by definition, triangle AEB is a right triangle.
This is further supported by Figure 3.4 that refers to the Pythagorean theorem, which is a fundamental component in establishing the properties of right triangles. In this case, it's not directly used to prove △AEB is a right triangle, but it's an underlying concept that supports the properties of the diagonals in a rhombus.