An isosceles triangle has two sides of equal length. In a rectangle, the diagonals are congruent and bisect each other, leading to the conclusion that triangle AEB, formed by one of the diagonals and two sides of the rectangle, is isosceles because AE equals BE.
To prove that triangle AEB is isosceles when ABCD is a rectangle and diagonals AC and BD intersect at E, we use the properties of a rectangle and the fact that diagonals of a rectangle are congruent.
Since ABCD is a rectangle, we know that AB is parallel to CD and AD is parallel to BC. Also, AC and BD bisect each other because they are diagonals of a rectangle.
Triangles AEB and DEC are formed by these intersecting diagonals.
In a rectangle, the diagonals are equal in length, therefore, AE is equal to CE, and BE is equal to DE.
Hence, triangles AEB and DEC are congruent by the Side-Side-Side postulate (SSS).
Since AE and BE are equal, triangle AEB must be an isosceles triangle with AE congruent to BE.
The isosceles triangle AEB is thus proven by the congruent segments AE and BE resulting from the congruent diagonals AC and BD of rectangle ABCD.
