As shown in the figure, the radius of the inscribed circle of RT △ ABC is 1 cm, and the hypotenuse is tangent to the circle O at point D. given AB = 5, find the length of AD and AC

Let ⊙ o cut AC to e, BC to F, quadrilateral OECF is square, ∵ CE = CF = 1, according to tangent length, we get AE = ad, BF = BD, ∵ AE + BF = AB = 5, let AE = x (x > 0), then BF = 5-x, ∵ AC = x + 1, BC = 6-x, according to Pythagorean theorem: (x + 1) ^ 2 + (6-x) ^ 2 = 25, 2x ^ 2-10x + 12 = 0x ^ 2-5x + 6 = 0

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1. As shown in the figure, BD is the diameter of ⊙ o, OA ⊥ ob, M is the point on the inferior arc AB, through point m as the tangent of ⊙ o, the extension line of OA at point P, MD and OA at point n. (1) verification: PM = PN; (2) if BD = 4, PA = 32ao, through point B as BC ∥ MP intersection ⊙ o at point C, the length of BC
2. As shown in the figure, BD is the diameter of ⊙ o, OA ⊥ ob, M is the point on the inferior arc AB, through point m as the tangent of ⊙ o, the extension line of OA at point P, MD and OA at point n. (1) verification: PM = PN; (2) if BD = 4, PA = 32ao, through point B as BC ∥ MP intersection ⊙ o at point C, the length of BC
3. As shown in the figure, the center of two concentric circles is O, the radius of big circle O is OA, OB intersects small circle O with C, D, please explain: AB / / CD I hope we don't use center angle and circle angle
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5. As shown in the figure, it is known that in △ AOB, ∠ AOB = 90 ° OD ⊥ AB is at point D. the circle with point o as the center and OD as the radius intersects OA at point E, intercept BC = ob on Ba, and prove that CE is the tangent of ⊙ o
6. As shown in the figure, ⊙ o radius OA is the diameter of ⊙ O1, ⊙ o another radius OC intersection ⊙ O1 and point B, try to explain the arc length relationship between chord AB and chord AC
7. As shown in the figure, OA, ob, OC are ⊙ o radius - AC = - BC, D, e are OA, ob, are the midpoint CD and CE equal? Why is it as shown in the figure
8. As shown in the figure, in RT △ AOB, the circle AB with radius OA intersects at point C. If Ao = 5, OB = 12, the length of BC is obtained
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10. As shown in the figure, in RT △ AOB, the circle AB with radius OA intersects at point C. If Ao = 5, OB = 12, the length of BC is obtained
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