Question

Show that the diagonals of a rhombus are perpendicular to each other.

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Answer 1

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Consider the rhombus as ABCD,

Let the center point be O

Now in triangle AOD and COD,

OA = OC ( Diagonals of IIgm bisect each other )

OD= OD (common )

AD = CD

Therefore, triangle AOD congruent triangle COD

Thus gives ,

Angle AOD = angle COD (cpct)

= 2 AOD = 180°

= AOD = 90°

So , the diagonals of a rhombus are perpendicular to each other.

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Hope it helps you :)

Answer 2

Given...

ABCD is a rhombus...

To prove...

The diagonals of a rhombus are perpendicular to each other...

Solution...

We know that ABCD is a rhombus...

Therefore,

AB = BC = CD = CA

Now,

• In ∆AOB & ∆COB

OA = 0C ----- ( diagonals of a parallelogram intersects each other )

OB = OB ----- ( Common )

AB = CB ----- ( Sides of rhombus are equal )

Hence, ∆AOB & ∆COB are congruent to each other by SSS property...

Therefore,

angle AOB = angle COB --- ( by CPCT )

So, all sides of a rhombus are perpendicular to each other...