Question

prove that diagonals of parallelogram cut at right angle it is a rhombus​

Answer 1

Step-by-step explanation:

Let ABCD is a rhombus.

⇒ AB=BC=CD=DA [ Adjacent sides are eqaul in rhombus ]

In △AOD and △COD

⇒ OA=OC [ Diagonals of rhombus bisect each other ]

⇒ OD=OD [ Common side ]

⇒ AD=CD

∴ △AOD≅△COD [ By SSS congruence rule ]

⇒ ∠AOD=∠COD [ CPCT ]

⇒ ∠AOD+∠COD=180

o

[ Linear pair ]

⇒ 2∠AOD=180

o

.

∴ ∠AOD=90

o

.

Hence, the diagonals of a rhombus bisect each other at right angle.

Answer 2

Answer:

First Proof : Since the diagonals AC and BD of quadrilateral ABCD bisect each other at right angles. Therefore, AC is the perpendicular bisector of the segment BD. ... Thus , ABCD is a quadrilateral whose diagonals bisect each other at right angles and all four sides are equal. Hence, ABCD is a rhombus.