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Answer:
The missing statement is ∠ACB ≅ ∠ECD
Step-by-step explanation:
Given two lines segment AC and BD bisect each other at C.
We have to prove that ΔACB ≅ ΔECD
In triangle ACB and ECD
AC=CE (Given)
BC=CD (Given)
Now to prove above two triangles congruent we need one more side or angle
so, as seen in options the angle ∠ACB ≅ ∠ECD due to vertically opposite angles
hence, the missing statement is ∠ACB ≅ ∠ECD
Answer:
SAS
Step-by-step explanation:
The two triangles share a side, so that would be reflexive to show that side is congruent. Also, keep in mind that all right angles are congruent! The picture already tells us that the two outer side are congruent. So in conclusion, the two angles are congruent by SAS.
Hope this helps! :)
Let's see
In ∆ABE and ∆CBE
- BE=BE(Common side)
- AE=EC[Diagonals of a parallelogram bisect each other]
- <AEB=<BEC[90°]
So by
SAS congruence the triangles are congruent
AB=BC
Fact:-
It's already given AC is perpendicular to BD
- It means diagonals are perpendicular to each other
According to general property of rhombus this parallelogram is also a rhombus.
So sides are equal hence AB =BC
