![](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 △ ABC, ab = AC, e is on AC, and the extension line of ad = AE, de intersects BC at point F
It is proved that: as shown in the figure, am ⊥ BC is made in M through a, ∵ AB = AC, ∵ BAC = 2 ⊥ BAM, ∵ ad = AE, ∵ d = ∠ AED, ∵ BAC = 2 ⊥ BAM = 2 ⊥ D, ∵ BAM = D, ∵ DF ∥ am, ≁ am ⊥ BC, ≁ DF ⊥ BC
In the triangle ABC, ab = AC, ad is perpendicular to point D, DF is perpendicular to point F, AE is perpendicular to point E, and DF intersects point M
There are several ways to solve this problem
Let ad intersect CF with n
Proband BD = CD = dn
Then angle DFN = 1 / 2 angle BDF
Prove again that angle fam = angle DFN, so angle fam = 1 / 2 angle BDF
Further prove that angle BDF = angle BDA, so angle fam = 1 / 2 angle BDA
If am is the bisector of angle bad, the conclusion can be drawn
As shown in the figure, in the isosceles △ ABC, ab = AC, ad is the height on the edge of BC, and points E and F are the points on the edge AB and AC respectively, and ef ‖ BC. (1) try to explain that △ AEF is an isosceles triangle; (2) try to compare the size relationship between de and DF, and explain the reason
(1) The reasons are as follows: ∵ ad is the height on the bottom edge of the isosceles triangle ABC, ∵ ad is the bisector of ∵ BAC, and ∵ AEF is the isosceles triangle
In diamond ABCD, ∠ ABC = 60 °, e is a point on diagonal AC, f is a point on the extension line of segment BC, and CF = AE connects be and ef
(1) If e is the midpoint of line AC, be = EF is proved
(2) If e is a point on the line segment AC or AC extension, and other conditions remain unchanged, what is the quantitative relationship between the line segments be and ef?
(1) Be = AE * root sign 3-angle EBC = 60 degree / 2 = 30 degree CF = AE BF = 3 * AE cosine theorem: the square of EF = the square of be + the square of BF - 2 * be * BF * cos30 degree = the square of 3 * AE + the square of 9 * AE - 2 * (root sign 3 * AE) * (root sign 3) / 2 = the square of 3 * AE to get EF = AE * root sign 3 = be, namely be = ef (2)
There is a point P in the equilateral triangle ABC. The length of the line connecting P and each vertex is pa equal to 3, Pb equal to x, PC equal to 4, and the angle APC equal to 150 degrees
X = 4, rotate the triangle APC about point a for 60 degrees to get the triangle ap'b, connect p'p, get the triangle ap'p is an equilateral triangle, and then get the triangle pp'b is a right triangle, according to the Pythagorean theorem get Pb = 4
It is known that △ ABC is an equilateral triangle, P is a point in the triangle, PA = 3, Pb = 4, PC = 5
Rotate △ ABP clockwise about point B for 60 ° to get △ BCQ, connect PQ, ∵ - PBQ = 60 °, BP = BQ, ∵ - bpq is equilateral triangle, ∵ PQ = Pb = 4, and PC = 5, CQ = 4. In △ PQC, pq2 + QC2 = PC2, ∵ PQC is right triangle, ∵ - BQC = 60 ° + 90 ° = 150 °, and ∵ - APB = 150 °
In △ ABC, ∠ ACB is the angle RT, AC = BC, P is a point in △ ABC, and PA = 1, Pb = 3, PC = 2. Can you find the degree of ∠ APC?
Take C as the vertex, rotate BC to AC, connect P, p-prime.then triangle pp-prime C is isosceles right triangle, and triangle pp-prime A is right triangle. Then angle APC = 90 + 45 = 135 degrees. The above is a simple proof. Xiangjie proves it by himself, which is helpful to improve
In the triangle ABC, BC = 1, angle ACB is right angle, and angle ABC = 60 degrees. P is a point in the triangle, so that angle APB = angle APC = angle BPC. Calculate PA, Pb, PC
Let PA = a, Pb = B, PC = C. by using cosine theorem, we can know that a ^ 2 + B ^ 2 + 1 / 2 * a * b * cos120 degree = a ^ 2 + B ^ 2 + AB = AB ^ 2 = 4, formula (1) a ^ 2 + C ^ 2 + 1 / 2 * a * c * cos120 degree = a ^ 2 + C ^ 2 + AC = AC ^ 2 = 3, formula (2) C ^ 2 + B ^ 2 + 1 / 2 * c * b * cos120 degree = C ^ 2 + B
It is known that in the arithmetic sequence {an}, the sum of the first 30 terms S30 = 50, the sum of the first 50 terms S50 = 30, then the sum of the first 80 terms is?
Finish in ten minutes, with score
S50-S30=a31+a32
+… +a50= [20*(a31+a50)]/2=10(a31+a50)=10(a31+a80)=30-50=-20.
∴a1+a80=-2
∴S80=[80*(a1+a80)/2]=-80
Answer - 80
In known arithmetic sequence an, the sum of the first 30 items is 50, and the sum of the first 50 items is 30, then the sum of the first 80 items is known to me
Given that the sum of the first 30 items in the arithmetic sequence an is 50, and the sum of the first 50 items is 30, then the sum of the first 80 items is the simpler algorithm I know, but which of the following algorithms is wrong? Why can't we calculate - 80? S30 = A1 × 30 + (30 × 29) / 2 × d = 50, S50 = A1 × 50 + (50 × 49) / 2 × d = 30, we get A1 = 3.6 d = - 4 / 30, then S80 = 3.6 × 80 + (80 × 79) / 2 × 4 / 30 =?
Because your explanation is wrong:
a1=241/75,d=8/75
RELATED INFORMATIONS
- 1. Three straight lines PA, Pb, PC, ∠ APC = ∠ APB = 60 ° and ∠ BPC = 90 ° in space are used to calculate the dihedral angle b-pa-c
- 2. The three sides of triangle are equal difference series, the circumference is 36, and the circumference of inscribed circle is 6 π Some people do that ∵ the three sides grow into an arithmetic sequence with a circumference of 36 Let the three sides be 12-n, 12, 12 + n respectively The circumference of the inscribed circle is 6 π --- radius of the inscribed circle r = 3 --- triangle area = (36 / 2) * 3 = 54 From Helen's formula: 54 & sup2; = 18 * 6 * (6 + n) * (6-n) --->27=36-n²--->n²=9--->n=3 --->The lengths of the three sides are 9, 12 and 15 respectively ∵ 9 & sup2; + 12 & sup2; = 15 & sup2;, ∵ the triangle is a right triangle Then why must one side be 12?
- 3. If the angle c = 90 degrees, (a + b) * (a + b) = 12 and C = 2 in the right triangle ABC, then the s triangle ABC =?
- 4. In the right triangle ABC, angle c is right angle, angle B ∶ angle c = 1 ∶ 4, find the degree of angle A The middle equation is still missing
- 5. In the right triangle ABC, C is equal to 90 degrees, AC is equal to twice BC, then the value of sina is?
- 6. In the triangle ABC, a = 120 °, ab = 5, BC = 7, then the value of sina / SINB
- 7. If the square of a plus the square of B equals twice the square of C, then the minimum value of COSC is
- 8. It is known that: as shown in the figure, the circle O with the diameter of AB on one side of the equilateral triangle ABC intersects with the sides AC and BC at points D and e respectively, DF ⊥ BC is made through point D, and F is the perpendicular foot (1) Proof: DF is tangent of circle O (2) If the side length of equilateral triangle ABC is 4, find the length of DF (3) Finding the area of the shadow in a graph The first question is not, the shadow area is DFE
- 9. In △ ABC, D and E are on the lines AB and AC respectively, ad = 2, BD = 5, EA = 4, EC = 10, Ed: BC=
- 10. It is known that CD is the height of AB side in △ ABC, AC + BC = 8, CD = 2, AE is the diameter of circumscribed circle of △ ABC, (1) try to explain △ CBD ∽ AEC, (2) let AC = x, AE = y, find the functional relation of y with respect to x, and (3) find the maximum value of Y
- 11. If alpha is equal to - 3, then which quadrant is the end of alpha?
- 12. In known △ ABC, a = 30 °, C = 45 °, a = √ 2, solve triangle There are two more questions, please master solution, online and so on. 2. It is known that in △ ABC, a = √ 2, B = 2, a = 30 ° to solve the triangle. 3. In △ ABC, a = 1, B = √ 3, B = 60 ° to solve the triangle.
- 13. As shown in the figure, ∠ ABC = ∠ CDB = 90 °, AC = a, BC = B, when BD and a, B meet what kind of relationship, △ ABC ∽ Δ CDB?
- 14. In the triangle ABC, ∠ BAC is the acute angle, h is the intersection of high AD and be, ad = BD, BH = AC
- 15. In the isosceles triangle ABC, when AB = AC rotates △ ABC clockwise about point B for 35 degrees, point a just falls on point D on BC, and point C falls on point E, then ∠ Dec=
- 16. As shown in the figure, in △ ABC, BD bisects ∠ ABC, de ⊥ AB in E, ab = 3cm, BC = 2.5cm, and the area of △ abd is 2cm2. Calculate the area of △ ABC
- 17. It is known that AB is the diameter of the circle O, C is the point on the circumference different from a and B, PA is perpendicular to the plane where the circle O is located, AE ⊥ PC is in E, find the plane Abe ⊥ plane PBC
- 18. If the edge of the regular triangle ABC is 15 and the distance from a point p outside the plane ABC to a, B and C is 20, find the distance from point P to plane ABC
- 19. There are three ABC points on the plane. Draw a straight line through each two points. You can draw a straight line at most There are three ABC points on the plane. Draw a straight line through each two points. You can draw () lines at most and () lines at least It takes at least () nails to fix the wood to the wall Why? The following statement is correct: there are two kinds of positional relations between () 1, a and line L 2. There are three intersections when three lines intersect
- 20. In the triangle ABC, ab = AC = 13, BC = 10, point D is the midpoint of BC, de ⊥ AB, and the perpendicular foot is point E, then what is de equal to?