Question

ABCD is a square, C is the centre of the circle. Prove that AP = AR.

Answer 1

Given:

• ABCD is a square.
• C is the center of the circle.
• P and R are points on the circle and we are to prove that AP = AR.

Proof:

1. Properties of the square:

• Since ABCD is a square, all sides of the square are equal, i.e., AB = BC = CD = DA. 
• The diagonals of a square are equal in length and bisect each other at right angles. Thus, the diagonals AC and BD are of equal length and they intersect at the center of the square, which is also the center of the circle since the circle’s center is given as C.

2. Center of the circle:

• The center of the circle is the same as the center of the square, C. This means that the radius of the circle is the distance from C to any point on the circle.

3. Symmetry of the square and circle:

• The square ABCD has four lines of symmetry the diagonals and the midlines of the sides. This symmetry ensures that any point on the circle that is a reflection of another point through the center will be equidistant from the center.

4. The position of points P and R:

• Points P and R lie on the circle and because of the symmetry of the square, they must be reflections of each other through the center of the square since the center of the square is also the center of the circle.

5. Equality of distances:

• Since the distance from the center of the square C to any point on the circle is the radius of the circle, the distances AP and AR are both equal to the radius of the circle.
• Therefore, AP = AR, as they are both radii of the same circle.

Thus, we have proven that AP = AR. The symmetry of the square and the fact that both points lie on the circle centered at C ensures that these distances are equal.