Given: ABCD is a rectangle, diagnoals AC and BD intersect at E
Prove: Triangle AEB is isosceles
To prove that triangle AEB is isosceles, we need to show that two sides of the triangle are congruent. In this case, we need to show that AE is congruent to BE.
Here's how you can prove it:
1. Start by considering triangle AEB.
2. We know that ABCD is a rectangle, which means opposite sides are parallel and congruent. So, AB is parallel and congruent to CD, and AD is parallel and congruent to BC.
3. Since ABCD is a rectangle, diagonals AC and BD bisect each other at point E. This means that AE is congruent to EC, and BE is congruent to ED.
4. Now, we have triangles AEB and CEB, where AE is congruent to EC, and BE is congruent to ED.
5. By using the substitution property of equality, we can substitute EC with AE and ED with BE in triangle CEB. So, we have triangle CEB where CE is congruent to AE, and DE is congruent to BE.
6. Now, if we compare triangles AEB and CEB, we have AE congruent to CE, and BE congruent to DE.
7. By using the transitive property of equality, we can say that AE is congruent to CE, and BE is congruent to DE, so AE is congruent to BE.
8. Therefore, we have shown that triangle AEB is isosceles because AE is congruent to BE.
By following these steps, you can prove that triangle AEB is isosceles given that ABCD is a rectangle and the diagonals AC and BD intersect at point E.
In a previous question you probably proved that in a rectangle the two diagonals are equal and bisect each other, making all 4 smaller segments equal to each other.
Now prove triangles AED and CEB congruent and each is isosceles, and from that angle DAE = angle CBE
therefore their complements are equal.
therefore the base angles of triangle ABE are equal making it isosceles.
Write it up using the method and procedure that your teacher or the course requires.