Which of the following explains how ΔAEB could be proven similar to ΔDEC using the AA similarity postulate?

Quadrilateral ABDC, in which point F is between points A and C, point G is between points B and D, point I is between points A and B, and point H is between points C and D. A segment connects points A and D, a segment connects points B and C, a segment connects points I and H, and a segment connects points F and G. Segments AD, BC, FG, and IH all intersect at point E.

beanburito

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

answer: this is not a question

step-by-step explanation: this is a answer