show that if the diagonal of a quadrilateral bisect each other at right angle, then it is a rhombus
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Step-by-step explanation:
The diagonals are AC and BD
Let the diagonals bisect each other at O.
In ΔAOBandΔAOD
OA=OA (common)
OB=OD (given the bisect)
∠AOB=∠AOD (each 90° )
∴ΔAOB≅ΔAOD (SAS criteria)
The corresponding parts are equal.
AB=AD
Similarly, AB=BC
BC=CD
CD=AD
∴AB=BC=CD=DA
i.e. the quadrilateral is a Rhombus
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