What is the answer of this

Question

Two congruent chords AB and CD which are not diameters, intersect at right angle in P. O is the centre of the circle. If M and N are the mid-points of AB and CD respectively, then prove that quadrilateral OMPN is a square.

Answer 1

Hi Laxman,

 

It looks like you have an image like this:

 

If horizontal chord is AB and vertical chord is CD, then it will be a square for figure OMPN.

 

From what I've observe, the diameter will be √2 times the leg.

 

 

 

 

From this perspective, midpoint of each leg is 0.5 or 1/2 of original length (1 or x).  And midpoint of diameter (hypotenuse) will be √2/2.  From here, you have to calculate the new lengths of each leg to see if they add up to half of diameter (hypotenuse) (a² + b² = c²) and to see if each leg are perpendicular or form a right angle.  Basically, you will have a square inside of 45-45-90 triangle.

 

(leg 1)²  +  (leg 2)²  = (diameter/2)²

    a²      +     b²       =      c²

 

(0.5)²    +     (0.5)²  =  (√2/2)²

0.25      +      0.25    =  (2/4) = 1/2

 

0.5 = 1/2  It works.

 

How about with an x?

 

(x/2)²  +  (x/2)²  =  (x√2/2)²

 

(x²/4)    +   (x²/4)   =  (2x²/4) = (x²/2)

2x² / 4   =  x² / 2

x² / 2  =  x² / 2   It works as well.

 

I hope you got it.