Let's start with a special case. Let c be the center of the circle. Draw a horizontal diameter from a to b, right through c. Then draw a radius up and to the right, from c to r. Then draw the chord from a to r. The isosceles triangle acr has equal base angles. Let x be the measure of the base angle. Now ∠acr is 180-2x, and the supplementary ∠bcr is 2x. Thus ∠bar is half of ∠bcr. In other words, the inscribed angle is half the central angle.
Add to this picture a radius down and to the right from c to q. Reasoning as above, ∠qab is half of ∠qcb. Add these together and ∠qar is half of ∠qcr. Once again the inscribed angle is half the central angle. This reasoning holds whenever c is inside ∠qar.
Let c lie outside ∠qar. In other words, q and r are both above the diameter ab. As above, ∠baq is half of ∠bcq, and ∠bar is half of ∠bcr. Subtract angles to show that ∠qar is half of ∠qcr. That completes the proof.
