Semicircle as a special case. Open split view. Set the left board to Theorem 1 and the right to Theorem 2. On the left, drag A and B so the centre angle is exactly 180° (a straight line through the centre). What is the angle at the circumference? Compare with the right board. Theorem 2 is what happens when Theorem 1’s centre angle becomes a straight line — explain why.

From “same segment” to cyclic quadrilateral. Set left to Theorem 3, right to Theorem 4. On the left, drag Q across the chord into the opposite segment. The four points A, P, B, Q now form a cyclic quadrilateral. Reveal the angles on the left — what do they sum to? Compare with the right. The “supplementary angles in opposite segments” you saw in Theorem 3 is Theorem 4.

Two tangents = Tangent-radius applied twice. Set left to Theorem 5, right to Theorem 7. On the right, notice the two right-angle squares at T₁ and T₂. Each one is Theorem 5! Theorem 7 is just “Theorem 5 applied at two different points.” Use this to explain why ET₁ = ET₂ (hint: triangles OT₁E and OT₂E share OE and have OT₁ = OT₂ = r, and both have a right angle — they’re congruent).

Alternate segment and angle at the centre. Set left to Theorem 6, right to Theorem 1. Look carefully: the tangent at T in Theorem 6 is the “limit” of a chord whose other endpoint moves to coincide with T. The angle between the tangent and chord TA is half the central angle subtended by TA. Verify by setting up matching configurations on both boards.

The hidden cyclic quadrilateral in two tangents. Set left to Theorem 4, right to Theorem 7. In Theorem 7, reveal the angles at O and E — they sum to 180°. The two right angles at T₁ and T₂ also sum to 180°. So OT₁ET₂ has both pairs of opposite angles summing to 180° — which means OT₁ET₂ is itself a cyclic quadrilateral! What circle is it inscribed in? (Hint: it’s the circle with diameter OE.)