Question

The figure below shows rectangle ABCD.

Answer 1

Answer: We can find out the missing statement with help of below explanation.

Step-by-step explanation:

We have a rectangle ABCD with diagonals AC and BD ( shown in given figure.)

We have to prove: Diagonals AC and BD bisect each other.

In triangles, AED and BEC.

$\angle ADB \cong \angle CBD$ ( By alternative angle theorem)

$AD\cong BC$ ( Because ABCD is a rectangle)

$\angle CAD\cong \angle ACB$  ( By alternative angle theorem)

By ASA postulate,$\triangle AED\cong \triangle BEC$

By CPCTC, $BE\cong ED$ and $CE\cong EA$

⇒ BE= ED and CE=EA

By the definition of bisector, AC and BD bisect each other.

Answer 2

Answer:

To Prove the diagonals of the rectangle bisect each other, Statement with reason is given:

Statement  ABCD is a rectangle.

Reason:Given

Statement

:Opposite sides are parallel.(AB║DC)

Reason: Definition of a Parallelogram

Statement

:Opposite sides are parallel.(AD║BC)

Reason: Definition of a Parallelogram

Statement

:∠CAB ≅ ∠ ACB

Reason: Alternate interior angles theorem

Statement

: ∠ADB ≅ ∠CBD

Reason:  Alternate interior angles theorem

Option D:∠CAB ≅ ∠ ACB

In ΔAED and ΔBEC

∠CAB ≅ ∠ ACB→→Alternate interior angle,as AB║DC.

∠ADB ≅ ∠CBD→→Alternate interior angle,as AD║BC.

AD=BC→→Opposite sides in a rectangle are equal.

ΔAED ≅ ΔBEC⇒⇒[ASA]

AE=EC→[CPCTC]

BE=ED→[CPCTC]