It is known that AB in semicircle o is the diameter ad = DC ∠ cab = 30 ° AC = 3 root sign 3, and the length of ad is calculated Don't use sin to solve the problem that we didn't learn
connect BC.BD Because angle DAB and angle DCB are two circular angles of the same chord (complementary) and ACB = 90 (circular angle of diameter) cab = 30. DAC = DCA (isosceles), so 2dac + BAC + ACB = 180. The solution is DAC = 30. Then it is compared with three sides of triangle ade
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- 1. AB is the diameter of semicircle o, C is the point on semicircle o which is different from a and B, CD ⊥ AB, perpendicular foot is D, ad = 2, CB = 4 * radical 3, then CD = -? The root number can't be typed = =, by the way, how to type the root number--
- 2. As shown in the figure, in △ ABC, ab = AC, the semicircle o with AC as the diameter intersects AB and BC at points D and e respectively. (1) prove that point E is the midpoint of BC; (2) if ∠ cod = 80 °, calculate the degree of ∠ bed
- 3. As shown in the figure: in △ ABC, the semicircle o with ab as the diameter intersects AC BC at the point de. prove that the triangle ode is an equilateral triangle
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- 6. As shown in the figure, take the waist ab of the isosceles triangle ABC as the diameter to draw a semicircle o, intersecting AC with E and BC with D Verification: 1. D is the midpoint of BC 2. If the angle BAC = 50 degrees, calculate the degree of arc BD
- 7. As shown in the figure, AB is a fixed length line segment, the center O is the midpoint of AB, AE and BF are tangent points, e and F are tangent points, satisfying AE = BF. Take point G on EF, and point G is the extension line of tangent intersection AE and BF at points D and C. when point G moves, let ad = y, BC = x, then the functional relationship between Y and X is () A. Positive proportion function y = KX (k is constant, K ≠ 0, X > 0) B. primary function y = KX + B (k, B is constant, KB ≠ 0, X > 0) C. inverse proportion function y = KX (k is constant, K ≠ 0, X > 0) d. quadratic function y = AX2 + BX + C (a, B, C is constant, a ≠ 0, X > 0)
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- 11. As shown in the figure, in the semicircle AOB, ad = DC, ∠ cab = 30 °, AC = 23, the length of ad is calculated
- 12. It is known that in the semicircle AOB, ad = DC, ∠ cab = 30 °, AC = 2 times root sign 3, finding the length of AD, etc For help, add 447341718 to the picture,
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