Question

If in a quadrilateral ABCD, AC⊥BD and AC and BD bisect each other, then prove that it is a rhombus.

Answer 1

Answer:

We have a quadrilateral ABCD such that the diagonals AC and BD bisect each other at right angles at O.

∴ In ∆AOB and ∆AOD, we have

AO = OA [Common]

OB = OD [O is the midpoint of BD]

∠AOB = ∠AOD [Each 90°]

∴ ∆AOB ≅ ∆AOD [By SAS congruency]

∴ AB = AD [By C.P.C.T.] …(1)

Similarly, AB = BC …(2)

BC = CD …(3)

CD = DA …(4)

∴ From (1), (2), (3) and (4), we have

AB = BC = CD = DA

Thus, a quadrilateral ABCD is a rhombus.

Alternatively : ABCD can be proved first a parallelogram then proving one pair of adjacent sides equal will results in rhombus

Answer 2

Answer:

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Step-by-step explanation:

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