As shown in the figure, in the two concentric circles with o as the center, the chord AB and CD of the big circle are equal, and AB and the small circle are tangent at point E

As shown in the figure, in the two concentric circles with o as the center, the chord AB and CD of the big circle are equal, and AB and the small circle are tangent at point E

Prove: as shown in the right figure, connect OE, make of ⊥ CD through o at f. ∵ AB and small ⊙ o tangent at point E, ∵ OE ⊥ AB, ∵ AB = CD, ∵ OE = of (the chord center distance of the same circle is equal), and ∵ CD is tangent to small ⊙ o

As shown in the figure, in the two concentric circles with o as the center, the chord AB and CD of the big circle are equal, and AB and the small circle are tangent at point E

Prove: as shown in the right figure, connect OE, make of ⊥ CD through o at f. ∵ AB and small ⊙ o tangent at point E, ∵ OE ⊥ AB, ∵ AB = CD, ∵ OE = of (the chord center distance of the same circle is equal), and ∵ CD is tangent to small ⊙ o

As shown in the figure, we know that point O is the common center of two concentric circles, and the chord ab of the big circle intersects the small circle at two points c and D. (1) prove: AC = BD; (2) if AB = 8, CD = 4, find the area of the ring

(1) Through the point o as OE ⊥ AB in E, AE = be, CE = De, ae-ce = be-de, AC = BD; (2) connect OA, OC, in RT △ AOE and RT △ OCE: oe2 = oa2-ae2, oe2 = oc2-ce2, ∧ oa2-ae2 = oc2-ce2, ∧ oa2-oc2 = ae2-ce2, ∧ ab = 8, CD = 4, ∧ AE = 4, CE = 2, ∧ oa2-oc2 = 12, ∧ the area of the ring is: π oa2 - π oc2 = π (oa2-oc2) = 12 π