Given two concentric circles, prove that all chords of the outer circle that touch the inner circle at a point are equal.

Steps to Solve

1. Identify the Problem Setup

We have two concentric circles with a common center ( O ). There are two chords ( AB ) and ( CD ) located in the larger circle. We are tasked with proving that these chords are equal in length, given certain angles and properties.

2. Analyze Given Angles

From the problem, we know:

• ( \angle OCB = 90^\circ )
• ( \angle OAB = \angle OAD )

This means that points ( A ) and ( B ) lie on one line perpendicular to ( OC ).

3. Use the Properties of the Circle

Since ( OA = OB ) (both are radii of the larger circle) and ( OC = OD ) (radii of the smaller circle), we can state that:

$$ OA = OB = r_1 \quad \text{and} \quad OC = OD = r_2 $$

4. Apply the Right Triangle Theorem

In triangle ( OCB ) and triangle ( ODA ):

• We have two right triangles, which share the hypotenuse and two corresponding sides.
• Thus, we can apply the Pythagorean theorem:

$$ OB^2 = OC^2 + CB^2 $$

$$ OA^2 = OD^2 + AD^2 $$

5. Establish the Relationship Between Lines

By showing that both triangles are congruent and have equal corresponding sides, we conclude:

$$ AB = CD $$

6. Conclude the Proof

Based on the properties of the angles and the equal distances of the radii, it follows that the chords ( AB ) and ( CD ) must be of equal length.