Question

show that if the diagonal of a quadrilateral bisect each other at right angle,then it is a rhombus​

Answer 1

Answer:

For a rhombus PQRS, all sides must be equal, opposite sides parallel, and diagonals must bisect each other as well as be at 90° to each other.

Step-by-step explanation:

Let the intersection of diagonals be at O.

Since all 4 angles at O are 90°,SO^2+OR^2=RS^2 and likewise for all 4 triangles formed by diagonals in quadrilateral.

Also SO=OQ,

and PO=OR.

Now we've to prove that all 4 sides PQ QR RS and SP are equal.

It is also given that all angles at O are 90°

So SO2+OR2=RS2

but OR=OP

So SO^2+OP^2=SR^2

=SP^2 also

Now since all 4 triangles are congruent,

angle SRO=angle OPQ

(both these angles are opposite to the equal sides SO and OQ in the 2 congruent triangles.

And because these 2 angles are equal as well as alternate we can say for sure that PQ is parallel to RS.

Likewise we can prove all pairs of opposite sides parallel and by Pythagoras theorem and because diagonals are bisected,it is established that opposite sides are equal.

But since all 4 triangles inside the quadrilateral (formed by diagonals) are congruent, we can safely say that adjacent sides of quadrilateral are also equal .

All properties of Rhombus fulfilled/satisfied.

Therefore Quadrilateral is a RHOMBUS.

PROVED(?)

QED . Hope it helps.

Answer 2

Step-by-step explanation:

Take quadrilateral ABCD , AC and BD are diagonals which intersect at O.

In △AOB and △AOD

DO=OB ∣ O is the midpoint

AO=AO ∣ Common side

∠AOB=∠AOD ∣ Right angle

So, △AOB≅△AOD

So, AB=AD

Similarly, AB=BC=CD=AD can be proved which means that ABCD is a rhombus.