Read the proof.
Given: AE 1 EC; BD IDC
Prove: AAEC ~ ABDC
Statement
Reason
1. AE 1 EC;BD1DC
1. given
2. ZAEC is a rt. 2; ZBDC is a 2. definition of
Irt. 2
perpendicular
3. ZAEC = ZBDC
3. all right angles are
congruent
14. ?
4. reflexive property
5. AAEC - ABDC
5. AA similarity theorem
What is the missing statement in step 4?
O ZACE = ZBGB
OZEAB = ZDBC
O ZEAC = EAC
O ZCBD = ZDBC

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:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11

hi how are you doing today? if you ever need just tell me

step-by-step explanation: