Given: ABCD is a parallelogram and AC bisects BD.
Prove: AB is congruent to BC.
1 answer:
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
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