Question

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

Answer 1

Answer:

Step-by-step explanation:

Consider a quadrilateral ABCD

In which AC and BD are diagonals bisecting each other at right angle

Let the intersecting point be O

Then AOB is a right triangle with angle AOB=90°

In triangle AOB,

AB *AB=AO*AO+OB*OB

Similarly all the triangles formed while intersecting will be like this

That's we get AB=BC=CD=DA

Which can only occur in a square or a rhombus

But the point to note is that we can't identify it is a square or a rhombus

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.