Question

3. Prove that a cyclic rhombus is a square.​

Answer 1

If anyone asks to me, that what is a cyclic quadrilateral, then I simply say that it is a quadrilateral, whose opposite angle are supplementary.

So, It's a cyclic rhombus, then Let it's angles be A, B, C , and D.

YOU HAVE TO DRAW A FAIR DIAGRAM TO UNDERSTAND.

So, 1. angle A = angle C, (because these are the opposite angles of a parallelogram.)

And also, 2. angle A+ angle C = 180° (because these are the opposite angle os a cyclic quadrilateral).

Then what happens when I replace the angle C with angle A, in the equation 1st,

angle A + angle A = 180°,

2* angle A = 180°,

Then, angle A = 90°.

Also, according to the first equation, angle C = 90°,

Finally, all the angles are equal to 90°.

But this answer only proves that the figure is a rectangle.

But wait...

It is a rhombus also, then my figure is a parallelogram with all side equal and angles of 90°.

That's why it's a square.

Answer 2

Step-by-step explanation:

Let ABCD be a cyclic rhombus.

$\therefore AB=BC=CD=AD.... (1)\\(Sides\: of\: rhombus)\\</p><p>\angle BAD= \angle BCD.... (2)\\ \&\\\angle ABC= \angle ADC.... (2)\\(Opposite \:angles\: of\: rhombus)\\</p><p>Now, \\</p><p>\angle BAD+ \angle BCD=180\degree.... (3)\\(Opposite\: angles \:of \:cyclic \:\square)\\ From \:equations \:(2) \: \& \:(3)\\</p><p>\angle BAD+ \angle BAD =180\degree\\</p><p>\therefore 2\angle BAD=180\degree\\</p><p>\therefore \angle BAD=\frac {180\degree}{2} \\</p><p>\therefore \angle BAD= {90\degree}\\</p><p>\therefore \angle BCD= {90\degree}\\</p><p>\boxed{\therefore \angle BAD=\angle BCD= {90\degree}}\\</p><p>Similarly, \:we \:can\: prove\: that:\\</p><p>\boxed{\angle ABC = \angle ADC = {90\degree}}\\</p><p>Thus, \\</p><p>{\tiny{\boxed{\angle ABC =\angle BCD=\angle ADC =\angle BAD= {90\degree}}}} ... (4)$

Hence, from equations(1) & (4) it can be observed that all sides are equal and each angle measures 90°.

Thus, it can be stated that a cyclic rhombus is a square.

Hence Proved.