If the diagonals AC and BD of a quadrilateral ABCD bisect each other, then ABCD is a :

parallelogram

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rectangle

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rhombus

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trapezium

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Solution

The correct options are
A parallelogram
B rectangle
C rhombus
The diagonals $A C$ and $B D$ of a quadrilateral $A B C D$ bisect each other, then $A B C D$ can be a parallelogram or any special type of parallelogram like rhombus, rectangle and square.
In a trapezium, the diagonals don't bisect each other.

Suggest Corrections

Similar questions

A C

B D

A B C D

A B C D

A B C D

(1) $A B C D$ is a Rhombus	(a) $¯ ¯¯¯¯¯¯ ¯ A C$ and $¯ ¯¯¯¯¯¯¯ ¯ B D$ bisect each other.
(2) $A B C D$ is a Parallelogram	(b) $¯ ¯¯¯¯¯¯ ¯ A C$ and $¯ ¯¯¯¯¯¯¯ ¯ B D$ bisect each other at an right angle.
(3) $A B C D$ is Rectangle	(c) $¯ ¯¯¯¯¯¯ ¯ A C$ and $¯ ¯¯¯¯¯¯¯ ¯ B D$ are congruent and bisect each other at an right angle.
(4) $A B C D$ is Square	(d) $¯ ¯¯¯¯¯¯ ¯ A C$ and $¯ ¯¯¯¯¯¯¯ ¯ B D$ are congruent and bisect each other.

Is quadrilateral ABCD a ||gm?

I. Diagonals AC and BD bisect each other.

II. Diagonals AC and BD arc equal.

The correct answer is: (a)/(b)/(c)/(d).

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