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As shown in the figure, ∠ C = 90 °, B = 60 ° in △ ABC, the point O is on AB, Ao = x, and the radius of ⊙ o is 1?
∵∠C=90°，∠B=60°，
∴∠A=30°，
∵AO=x，
∴OD=1
2AO=1
2x，
(1) If the circle O is separated from AC, then od is greater than R, i.e. 1
2X > 1, x > 2;
(2) If the circle O is tangent to AC, then od is equal to R, i.e. 1
2X = 1, x = 2;
(3) If the circle O intersects AC, then od is less than R, i.e. 1
2X < 1, 0 < x < 2;
To sum up, when x > 2, AC is separated from ⊙ o; when x = 2, AC is tangent to ⊙ o; when 0 < x < 2, AC intersects ⊙ o
In the RT triangle ABC, ∠ a = 90 °, C = 60 °, O moves on BC, Bo = x, radius of circle O is 2 When x is in what range, AB intersects, tangents, and separates from circle O?
Let oh ⊥ AB in H, let the distance between O and ab be D, then Oh = D is known, ﹤ B = 90 ° - 60 ° = 30 ° in RT △ boh, oh = Bo / 2, i.e. d = x / 2. When d = x / 2 < R = 2, i.e. x < 4, AB intersects with circle O; when d = x / 2 = r = 2, i.e., x = 4, AB is tangent to circle O; when d = x / 2 > R =..., AB is tangent to circle o
As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 5, BC = 12, ⊙ o radius is 3 (1) If the center O and C coincide, what is the position relationship between ⊙ O and ab? (2) If the point O moves along the ray Ca, what is the ⊙ o tangent to ab when OC is equal to?
(1) According to the Pythagorean theorem, we get that: ab = ac2 + BC2 = 52 + 122 = 13. From the area formula of triangle, we can get: AC × BC = ab × CD, 5 × 12 = 13 × CD,
As shown in the figure, in RT △ ABC, ∠ C = 90 ° BC = 5, ⊙ O and the three sides AB, BC and AC of RT ⊙ ABC are tangent to points D, e and f respectively. If the radius of ⊙ o is r = 2, then the circumference of RT ⊙ ABC is______ .
Connect OE, of, let ad = x, from the tangent length theorem, we get AF = x, ∵ O and RT △ ABC, whose three sides AB, BC and AC are tangent to points D, e, F, ⊥ OE ⊥ BC, of ⊥ AC, ⊥ the quadrilateral OECF is a square, ∵ r = 2, BC = 5, CE = CF = 2, BD = be = 3, ? obtained from Pythagorean theorem, (x + 2) 2 + 52 = (x + 3) 2
As shown in the figure of triangle ABC, the angle B is equal to 90 degrees, the angle c is equal to 60 degrees, the point O is on AC, Ao is equal to x, and the radius of circle O is equal to X. when x is taken in what range Tangent of O, O and ab?
As OD and CB are both perpendicular to AB, and AOD is 60 degrees, then od = 1 / 2ao = 1 / 2x
Od is the distance from the center of the circle to AB, and 1 / 2x, so the line AB intersects the circle o
The triangle ABC can form three triangles when there is one point on BC side (point D is connected with vertex a), and six triangles can be formed when there are two points on BC edge When there are three points on BC side, 10 triangles can be formed. When there are 0 points on BC side, one triangle can be formed. Question: how many different triangles can be formed when there are n points on BC side (not coincident with B and C)?
（N+1）（N+2）/2
The three vertices of the triangle ABC are a (- 3,0), B (2,1), C (- 2,3) 1. Find the equation of the straight line where BC is located. 2. The equation of straight line where the center line ad on the side of BC is located. 3. The equation of vertical bisector de on BC side
(1) The slope of the line where the BC side is located is (3-1) / (- 2-2) = - 1 / 2. Therefore, the linear equation is: Y-1 = (- 1 / 2) (X-2) (point oblique) that is, x + 2y-4 = 0 (2) the slope of the line where ad is located: (2-0) / (0 + 3) = 2 / 3. Therefore, the equation of the line where ad is located is: y = (2 / 3) (x + 3), that is, 2x-3y + 6 = 0 (3) of the line on which BC is located
As shown in the figure, the circumference of the triangle ABC is 32, and the midpoint of its three sides is taken as the vertex to form the second triangle, The third triangle is formed by taking the midpoint of the three sides of the second triangle as the vertex
If the circumference should be divided by 2 in turn, then the circumference of the nth triangle is the N-1 power of 32 △ 2
As shown in the figure, the circumference of △ ABC is 32. The second triangle is formed by taking the midpoint of its three sides as the vertex, and then the third triangle is formed by taking the midpoint of the three sides of the second triangle as the vertex Then the circumference of the nth triangle is______ .
According to the theorem of the median line of a triangle, the length of each side of the second triangle is equal to half of the sides of the largest triangle,
Then the circumference of the second triangle is equal to the circumference of △ ABC × 1
2=32×1
2，
The circumference of the third triangle is = △ ABC perimeter × 1
2×1
2=32×（1
2）2，
...
Circumference of the nth triangle = 32 × (1
2）n-1=26-n，
So the answer is: 26-n
As shown in the figure, △ ABC is inscribed in ⊙ o, the chord cm ⊥ AB is in M, CN is the diameter, f is the diameter The midpoint of AB, Verification: CF bisection ∠ MCN
Proof: connect of,
∵ f is
The midpoint of AB,
Of bisect ab
∴OF⊥AB．
And ∵ cm ⊥ ab,
∴CM∥OF．
∴∠MCF=∠OFC．
And ∵ OC = of,
∴∠OCF=∠OFC．
∴∠MCF=∠OCF．
ν CF bisection ∠ MCN
RELATED INFORMATIONS
- 1. Circle O is the circumscribed circle of triangle ABC, chord cm is perpendicular to AB, CN is the diameter, f is the midpoint of arc ab. the bisection angle NCM of CF is proved
- 2. The vertex C passing through △ ABC as a straight line intersects with edge AB and median line ad at points F and e respectively. It is proved that AE / de = 2AF / BF
- 3. O is a point in the equilateral triangle ABC, and the over current o is OD ∥ AB intersecting BC at point D and OE ∥ AC crossing BC at point E, what triangle ode is?
- 4. (1) As shown in Figure 1, ab = BC = Ca, OD and OE are radius of ⊙ o, OD ⊥ BC is at point F, OE ⊥ AC is at point G, It is proved that the area of the shadow quadrilateral ofcg is 1 of the area of △ ABC 3． (2) As shown in Fig. 2, if ∠ DOE keeps 120 ° angle unchanged, It is proved that when ∠ DOE rotates around point O, the area of the figure (shadow part in the figure) surrounded by two radii and two edges of △ ABC is always 1 of the area of △ ABC 3．
- 5. It is known that in △ ABC, D is the midpoint of AB, e is a point on AC, AE = 2ce, CD, be intersect at o point, OE = 2cm. Find the length of Bo
- 6. In the triangle ABC, ab = AC, and the midpoint of BC is d. draw an equilateral triangle def so that the vertices E and F are parallel to BC on sides AB and AC respectively
- 7. As shown in the figure, in △ ABC, ∠ BAC = 90 °, ab = AC, point D is on BC, and BD = Ba, point E is on the extension line of BC, and CE = Ca, try to find the degree of ∠ DAE (1) If the condition of "ab = AC" is removed and the other conditions remain unchanged, will the degree of ∠ DAE change?
- 8. As shown in the figure, in the triangle ABC, the angle c = 90 degrees, ad is the bisector of angle a, ab = 26, AC = 10cm, and find the distance from D to ab From top to bottom, from left to right is ABCD
- 9. In RT △ ABC, ∠ C = 90 degrees, square defg is connected to △ ABC, De is on AB, F and G are on BC and AC ~ B respectively In RT △ ABC, ∠ C = 90 degrees, square defg is connected to △ ABC, De is on AB, F, G are on BC and AC respectively, BC = 3, AC = 4, find ad: de: EB
- 10. In RT △ ABC, ∠ C = 90 °, AC = 3cm, BC = 4cm, then the distance from its outer center to the vertex of the right angle is______ cm．
- 11. If BC = 6, Ao is equal to the root sign 3, then the circle O is half
- 12. As shown in the figure, in △ ABC, ∠ C = 90 °, CA = CB, CD ⊥ AB in D, CE bisection ∠ BCD intersecting AB in E, AF bisection ∠ a intersecting CD in F. verification: EF is parallel to BC
- 13. As shown in the figure, in △ ABC, point E is on AB, point D is on BC, BD = be, ∠ bad = ∠ BCE, ad and CE intersect at point F. try to judge the shape of △ AFC and explain the reasons
- 14. Given that B and C are two fixed points, the absolute value of BC is equal to 8, and the circumference of triangle ABC is equal to 18, the trajectory equation of vertex A is obtained Why is there only one trajectory equation? Isn't there two cases? It is known that F1 and F2 are two focal points of the ellipse, and the point m satisfying the vector MF1 * MF2 = 0 is always inside the ellipse, then the eccentricity range of the ellipse is Why does the interior of the ellipse include boundaries?
- 15. The coordinates of the vertex B and C of the triangle ABC are (0,0) and (4,0) AB, and the length of the midline on the edge of (4,0) AB is 3
- 16. It is known that in the isosceles triangle ABC, ab = AC, ad ⊥ BC at D, CG / / AB, BG intersect ad, AC in E and f respectively, and it is proved that be 2 = EF × eg
- 17. In △ ABC, if high AD and be intersect at H point and BH = AC, then ∠ ABC=______ .
- 18. In △ ABC, AC = 10, ab = 17, BC edge height ad = 8, calculate △ ABC area (to draw a graph)
- 19. As shown in the figure, the triangle ABC is a right triangle. The area of shadow 1 is 23cm square. Find the length of BC
- 20. As shown in the figure, in △ ABC, ab = AC, ad is the angular bisector of ∠ BAC, ad = 8cm, BC = 6cm, and points E and F are two points on ad, then the area of shadow part in the figure is () A. 48 B. 24 C. 12 D. 6