Question

The diagonals of quadrilateral ABCD intersect at P. If ABparallel CD and AB≌ CD , which additional statement shows that ABCD is a rhombus?

A AP=CP
D BC=AD
C △ DPA≌ △ DPC
D BC parallel AD

Answer 1

To prove that quadrilateral ABCD is a rhombus, the statement BC=AD would directly confirm that all four sides are equal, making it the most conclusive additional proof.

Step-by-step explanation:

The question involves determining which additional statement shows that quadrilateral ABCD is a rhombus, given that AB is parallel to CD and they are of equal length. A rhombus is a special type of parallelogram where all four sides are of equal length. So the question demands an additional property that will confirm the equal length of all sides.

To unequivocally prove ABCD is a rhombus, we would need to show that one of the following is true:

• AP=CP, which would imply that the diagonals bisect each other at right angles. In any rhombus, the diagonals bisect each other and are perpendicular.
• BC=AD, confirming that opposite sides are equal, hence all four sides are equal.
• △ DPA ≅ △ DPC, proving that triangles formed by the diagonals are congruent, which would also imply that the diagonals bisect each other at right angles.
• BC parallel AD, which alone would not confirm that ABCD is a rhombus as it does not provide information about the equality of all sides of the bisecting property of the diagonals.

In conclusion, the additional statement that BC is equal to AD would be the most straightforward and conclusive proof that quadrilateral ABCD is indeed a rhombus.