Clara writes the following proof for the theorem: if the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram: clara's proof 1. for triangles aob and cod, angle 1 is equal to angle 2 as they are vertical angles. 2. ao = oc and bo = od because it is given that diagonals bisect each other. 3. the triangles aob and cod are congruent by sas postulate. 4. similarly, triangles aod and cob are congruent. 5. by angle abd is equal to angle cdb and angle adb is equal to angle cbd. 6. as the alternate interior angles are congruent the opposite sides of quadrilateral abcd are parallel. 7. therefore, abcd is a parallelogram.

a = 11 : 4

b = 38

c = 1 for every 6

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

13.46

step-by-step explanation:

16.80 x .20 = 13.46