Question 1

Let's use the Baudhāyana–Pythagoras theorem to prove that equal chords are equidistant from the circle's center.

Step 1 — Set up the diagram and knowns

Let's consider a circle with center O. We draw two chords, AB and CD. We are given that AB and CD have the same length. Let's draw a perpendicular from O to AB, meeting at M. Let's draw a perpendicular from O to CD, meeting at N. We know that a perpendicular from the center bisects the chord. So, M is the midpoint of AB, and N is the midpoint of CD.

$AM = \frac{AB}{2}$

$CN = \frac{CD}{2}$

We know that AB = CD. Therefore, their halves must also be equal.

$\boxed{AM = CN}$

Also, OA and OC are radii of the same circle.

$\boxed{OA = OC}$

Step 2 — Apply the Baudhāyana–Pythagoras theorem

Now, let's look at the right-angled triangle OMA. The Baudhāyana–Pythagoras theorem states that $hypotenuse^2 = side1^2 + side2^2$. Here, OA is the hypotenuse.

$OA^2 = OM^2 + AM^2$

Next, let's look at the right-angled triangle ONC. Here, OC is the hypotenuse.

$OC^2 = ON^2 + CN^2$

From Step 1, we know that OA = OC. So, their squares must also be equal.

$OA^2 = OC^2$

Now, we can substitute the expressions from the Baudhāyana–Pythagoras theorem.

$OM^2 + AM^2 = ON^2 + CN^2$

From Step 1, we also know that AM = CN. So, we can replace $CN^2$ with $AM^2$.

$OM^2 + AM^2 = ON^2 + AM^2$

Let's subtract $AM^2$ from both sides of the equation.

$OM^2 = ON^2$

Taking the square root of both sides, we get:

$\boxed{OM = ON}$

Answer

Since OM and ON represent the distances of the chords AB and CD from the center O, and we have shown that OM = ON, it proves that chords of a circle having the same length are at the same distance from the center.