![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
As shown in the figure, in the RT triangle ABC, the angle ACB is equal to 90 degrees, and ab is equal to 6. Take AC and BC as the diameters to make semicircles, and the areas are recorded as S1 and S2 respectively, then what is S1 plus S2 equal to
The answer is S1 + S2 = 9 π / 2
As shown in the figure, the diameter of circle O is ab = 4, ∠ ABC = 30 °, BC = 4 times the root sign 3, and D is the midpoint of line BC. Find the position relationship between D and circle O, and prove that 30° Urgent, urgent, urgent, urgent!
I don't know if this is the problem
(1) Try to judge the position relationship between point D and circle O, and explain the reasons. (2) pass through point D as de ⊥ AC, perpendicular foot is e, prove that De is tangent of circle o
If you agree with my answer, please click "adopt as satisfactory answer" and wish you progress!
AB is the diameter of ⊙ o, point C is on ⊙ o, BP is the center line of ⊙ ABC, BC = 3, AC = 6 2. Find the length of BP
∵ AB is the diameter of ⊙ o,
∴∠C=90°，
And ∵ BP is the center line of ∵ ABC,
∴CP=1
2AC=3
2，
In the right angle △ BCP, BP=
PC2+BC2=
(3
2)2+32=3
3．
As shown in the figure, in RT △ ABC, ∠ C = 90 ° AC= 2, BC = 1, if a circle with C as its center and CB as its radius intersects AB at point P, then AP=______ .
In RT △ ABC, ∵ C = 90 ° AC=
2，BC=1，
∴AB=
AC2+BC2=
3，
Let AC intersect m and extend AC to n,
Then am = ac-cm=
2-1 AN=
2+1
According to am? An = AP? AB,
(
2-1）（
2+1）=AP×
3，
AP is obtained=
Three
3．
It is known that in RT △ ABC, ab = radical, 2Ac = 1, take a as the center of the circle, AC as the radius, draw a circle crossing BC at the point P, and draw the long graph of BP by yourself
As ad vertical BC, then
AD*AD+CD*CD=AC*AC=1
Ad / CD = BD / ad
CD*BD+CD*CD=CD*(BD+CD)=CD*CB=1
Because BC = radical 3
CD = 1 / root 3, DP = CD = 1 / root 3
BP = bc-cp = root 3-2 / root 3 = root 3 / 3
As shown in the figure, the area of the equilateral triangle ABC is 3 pieces of 3cm? And the circle with a as the center and the line L where BC is located (1) There is no common point (2) has a unique common point (3) has two common points
The area of equilateral △ ABC is 3cm2
Let the side length of the equilateral △ ABC be a, then the area s = root 3 * a 2 / 4
So a 2 = 12
A = 2 root sign 3
The distance from a to BC d = radical 3 * A / 2 = 3
therefore
(1) When there is no common point, R
As shown in the figure, in RT △ ABC, ∠ C = 90 ° AC = 2, BC = 1, if a circle with C as its center and CB length as its radius intersects AB at point P, then the length of AP is () A. Three B. Three Three C. 2 Three Three D. 3
As shown in the figure, extend AC intersection ⊙ C to e, and let another intersection point with the circle be Q,
In RT △ ABC, ∠ C = 90 °, ∵ AC =
2，BC=1，
∴AB=
AC2+BC2=
3，
∵ CQ, CB, CE are the radius of a circle,
∴CQ=CB=CE=1，
According to the secant theorem, AQ · AE = AP · ab,
∴AP=AQ•AE
AB=(
2−1)(
2+1)
3=
Three
3．
Therefore, B
As shown in the figure, △ ABC, ab = AC, O is a point in △ ABC, and ∠ OBC = ∠ OCB
It is proved that: ∵ AB = AC, ABC = ∠ ACB (equilateral and equal angle),
∵∠OBC=∠OCB，
Ψ ABO = ∠ ACO, OB = OC (equiangular to equilateral),
∴△AOB≌△AOC（SAS），
∴∠OAB=∠OAC，
AC = ab,
⊥ BC
It is known that: as shown in the figure, the central line BD and CE of the triangle ABC intersect at the point O, F and G are the midpoint of OB and OC respectively (1) guess what kind of quantitative relationship and bit between EF and DG? It is known that: as shown in the figure, the central line BD and CE of the triangle ABC intersect at the point O, F and G are the midpoint of OB and OC respectively. (1) guess the quantitative and positional relationship between EF and DG (2) Prove your conjecture
EF is parallel to DG and EF = DG
Proof: connect Ao
If f is the midpoint of OB and E is the midpoint of AB, then EF is the median line of the triangle OAB, so EF is parallel to OA and equal to half of OA. Similarly, if DG is parallel to OA and equal to half of OA, EF is parallel to DG and EF = DG
As shown in the figure, in △ ABC, ab = AC, the two midlines BD and CE of △ ABC intersect at point o, Confirmation: OB = OC
It is proved that: two midlines BD and CE of ∵ △ ABC,
∴CD=1
2AC，BE=1
2AB，
∵AB=AC，
∴CD=BE，∠EBC=∠DCB，
In △ EBC and △ DCB
BE＝CD
∠EBC＝∠DCB
BC＝BC
∴△EBC≌△DCB（SAS），
∴∠DBC=∠ECB，
∴OB=OC．
RELATED INFORMATIONS
- 1. As shown in the figure, in the triangle ABC, the midlines BD and CE intersect at the points o, e and F are the midpoint of OB and OC respectively, indicating that the quadrilateral defg is a parallelogram
- 2. As shown in the figure, △ ABC is a right triangle, ∠ C = 90 °, CA = CB, ad bisect ∠ BAC, de ⊥ AB, if AB = 18cm, calculate the circumference of △ DBE
- 3. In the isosceles right triangle ABC, AC = BC, ∠ C = 90 degrees, ad bisects ∠ BAC, De is perpendicular to the point e AB = 15 cm, and the circumference of the triangle DBE is calculated In the isosceles right triangle ABC, the angle c is 90 degrees, AC = BC, ad bisection angle BAC, de vertical AB, perpendicular foot e, and ab = 15 cm?
- 4. In RT △ ABC, ∠ C = 90 ° cab = 60 ° and the distance between D and ab is 3.8cm, then how many centimeters is BC The properties of detailed points, especially the right triangle with 30 degree angle
- 5. As shown in the figure, BC is the chord of circle O, OD ⊥ BC is at point E, intersection arc BC is at point D, and point a is a moving point on the major arc BMC (not coincident with B and C). BC = 4, radical 3, ed = 2 are known (1) Find the radius of circle O (2) Find the degree of ∠ a (3) Find the position of point a when the shadow part takes the maximum value I can't figure it out
- 6. As shown in Figure 1, in triangle ABC, OB and OC are bisectors of angle ABC and angle ACB respectively. If angle a = x degrees, what is the degree of angle BOC
- 7. It is known that: point I is the inner heart of the triangle ABC, point O is the outer center of the triangle ABC, ∠ BOC = 140 ° to find ∠ BIC? Why is ∠ a = 70 ° or 110 ° more detailed?
- 8. 0 is a point in the triangle ABC, and the distance from O to the three sides is equal. Given the angle a = 70 °, calculate the degree of BOC
- 9. As shown in the figure, in △ ABC, ∠ B = 90 ° o is a point on AB, and a circle with o as its center and ob as its radius intersects with point e at point E
- 10. It is known that in △ ABC, ab = AC, BC = BD, ad = de = EB, then the degree of ∠ A is () A. 30° B. 36° C. 45° D. 50°
- 11. It is known as follows: in △ ABC, ∠ cab = 120 °, ab = 4, AC = 2, ad ⊥ BC, D are perpendicular feet. Find the length of AD
- 12. Known; a = 201x + 2011, B = 2010x + 2012, C = 2010x + 2013. Find a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ca
- 13. If the three sides of the triangle ABC are a, B, C, and satisfy the condition that the cubic power of a - the square of a B + the square of ab - the square of AC + the cubic power of BC - B = 0, find the shape of the triangle ABC
- 14. Known as | a || a+|b| b+|c| C = - 1, try to find ab |ab|+bc |bc|+ca |ca|+abc |The value of ABC |
- 15. Given that a = 2013 parts of 2012, B = 2012 parts of 2012, C = 1, find the algebraic formula 3 (AB + BC) minus 3 (AB minus AC) minus 4ac minus 3bC
- 16. Known: a + x ^ 2 = 2000, B + x ^ 2 = 2001, C + x ^ 2 = 2002, and ABC = 2, find the value of a / BC + C / AB + B / ac-1 / A-1 / B-1 / C
- 17. P is any point in the triangle ABC, which is proved; ab + BC + Ca > AP + BP + CP
- 18. Let P and Q be two points on the edge BC of △ ABC, and BP = PQ = QC = AP = AQ; find the size of angle BAC
- 19. Let p be a point in the plane of triangle ABC, and vector AP + 2 vector BP + 3 vector CP = vector 0, If vector AB = vector a, vector AC = vector B. (1) use vector a and vector B to represent vector AP and vector ad, (2) fill in the blank s triangle according to the above results PAB:S triangle PBC:S Triangle PAC = --- - -:----
- 20. It is known that the circle x ^ 2 + y ^ 2 + 2x-6y + F = 0 and X + 2y-5 = 0 intersect at two points a and B, and O is the origin of coordinates. If OA is perpendicular to ob, then the value of F is? The answer is zero·