The figure below shows rectangle abcd: the following two-column proof with missing statement proves that the diagonals of the rectangle bisect each other: statement reason abcd is a rectangle. given line segment ab and line segment cd are parallel definition of a parallelogram line segment ad and line segment bc are parallel definition of a parallelogram ∠cad ≅ ∠acb alternate interior angles theorem line segment bc is congruent to line segment ad definition of a parallelogram alternate interior angles theorem ƒ δ ade ≅ ƒ δ cbe angle-side-angle (asa) postulate line segment be is congruent to line segment de cpctc line segment ae is congruent to line segment ce cpctc line segment ac bisects line segment bd definition of a bisector which statement can be used to fill in the blank space? ∠adb ≅ ∠cbd ∠abe ≅ ∠ade ∠acd ≅ ∠ace ∠ace ≅ ∠cbd

Given : A  rectangle A B CD

To Prove: Diagonals of the rectangle bisect each other

Proof:

1. ABCD is a rectangle.  

→AB ║CD→ Definition of a Parallelogram

→AD║BC→ Definition of a Parallelogram

⇒∠CAD ≅ ∠ACB  →→[Alternate interior angles theorem]

⇒Line segment BC ≅ Line segment DA→→Definition of a Parallelogram

In  Δ A DE and Δ C BE

AD=BC⇒Proved above

∠CAD=∠ACB ⇒Alternate interior angles theorem

∠ADB=∠CBE ⇒Alternate interior angles theorem

→→Δ A DE ≅  Δ C BE⇒Angle-Side-Angle (A S A) Postulate

BE=DE→→[C P CT ]

A E=CE→→[C P CT ]

⇒Line segment AC bisects Line segment B D⇒Definition of a bisector

Blank Space : Option A⇒∠ADB ≅ ∠CBD

∠ADB ≅ ∠CBD

Step-by-step explanation:

its right bc i tried it

were is the equation

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.