Wow, that's a really interesting question. Do you mean you are wondering why the perpendicular from O naturally bisects the chord AB? One way to see this is to suppose that the perpendicular from O doesn't bisect the chord AB, and see that something doesn't work.

Let's start with our chord, and draw radii from the endpoints like this:

d358f4fa-848a-4125-9552-79367f6cabb2-image.png

The triangle ABO (where O is the center) has two sides the same length (it's isosceles), so the angles BAO and ABO are equal:

3d8ff75e-110e-45e8-a7d0-75b1d3725c3a-image.png

We want the line from O to the chord to be perpendicular to the segment AB. Let's suppose that it doesn't bisect the chord, so AD is not the same length as BD:

fd4b231b-9670-4840-b1bc-5e696aff2866-image.png

But look! The triangles ADC and BDC both have a right angle and angle a. This means that their remaining angles, ACD and BCD, must be equal. This means that the two triangles are similar! But their hypotenuses are both radii, so they must actually have sides that are the same length; in other words, be congruent. This means that AD and BD, which are corresponding legs on the two triangles, must be the same length!

This shows that the segment from the center O perpendicular to any chord must bisect the chord. This is true no matter where the chord is or how long it is!

Thanks for asking that really interesting question, and I hope that you are enjoying the course so far! There is nothing better than being able to prove a theorem or identity yourself. Great job with thinking!

Happy Learning,

The Daily Challenge Team