![](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>)
BD bisects ∠ ABC, be bisects ∠ ABC into two parts of 3:4, ∠ DBE = 8 ° and calculates the degree of ∠ ABC
Powder Z memories,
∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵
If ∠ Abe = 3A °, then ∠ CBE = 4A °
∴∠ABC＝7a°
∵ BD bisection ∠ ABC
∴∠ABD＝∠CBD＝3.5a°
∵∠DBE＝∠ABD－∠ABE＝3.5a°－3a°＝0.5a°＝8°
∴a＝16
∴∠ABC＝7a＝112°
BD bisects ∠ ABC, be divides ∠ ABC into two parts of 3:4, ∠ DBE = 8 degrees, and calculates the degree of ∠ ABC
=-=
Angle Abe = 3 / 7 angle ABC
Angle abd = 1 / 2 angle ABC
So angle EBD = 1 / 2-3 / 7 = 1 / 14, angle ABC = 8
So the angle ABC = 14 * 8 = 112 degrees
As shown in the figure, in △ ABC, ab = AC, BD bisects ∠ ABC, ad is the bisector of the outer angle ∠ EAC, connecting CD, trying to explain that △ ACD is an isosceles triangle
AB = AC, BD bisecting ∠ ABC. Angle abd = CBD = 1 / 2acb and ad is the bisector of outer angle EAC. Angle ead = DAB = 1 / 2 (ABC + ACB), DAC = ACB and AD / / BC are obtained. Therefore, ADB = DBC, abd = CBD, ADB = abd, ad = AB and ab = AC are obtained. Therefore, ad = AC is an isosceles triangle
The distance from the point of the bisector of the three inner angles of △ ABC to the three sides is equal to 2cm, and the perimeter of △ ABC is 18cm. What is the starting area
1. The distance from the point of the bisector of the three inner angles of △ ABC to the three sides is equal to 2cm, and the perimeter of △ ABC is 18cm
That's 18
Let the distance from the point of the bisector of the three inner angles of △ ABC to the three sides be h, and the three sides of △ ABC are respectively a, B, C
Then the area s = 0.5ah + 0.5bh + 0.5ch = 0.5 (a + B + C) H = 0.5 × 18 × 2 = 18 square cm
The distance between the points of the bisectors of the three inner angles of △ ABC and the three sides is equal to 2cm. The perimeter of △ ABC is 18cm. What is the area of △ ABC
Cross the intersection of the bisector, make the three sides of the vertical, AE, AF, Ag
AE=AF=AG
S=1/2AE*AB+1/2AF*AC+1/2AG*BC
=1/2(AE*AB+AF*AC+AG*BC)
=1/2(AE*(AB+AC+BC))
=1 / 2ae * perimeter
=1/2*2*18
=18
If the area and perimeter of the triangle ABC are 15 and 15, then the distance from the intersection of bisectors of each inner angle to each side is 0
The intersection of the bisectors of the internal angles is the inner part of the triangle, and the distance from the inner part to each side is the radius of the inscribed circle, R
The large triangle is divided into three equal height triangles with the radius r of the inscribed circle as high;
s=1/2*a*r+1/2*b*r+1/2*c*r
2s=(a+b+c)*r
30=15*r
r=2
If the two vertex coordinates a (- 4,0), B (4,0) of △ ABC and the perimeter of △ ABC are 18, then the trajectory equation of vertex C is___ ．
(1) A (- 4,0), B (4,0), perimeter 18, ab = 8, BC + AC = 10, ∵ 10 ＞ 8, the sum of distances from C to two fixed points is equal to the fixed value, the trajectory of C is an ellipse with a and B as the focus, ∵ 2A = 10, 2C = 8, ∵ B = 3, so the standard equation of ellipse is X225 + Y29 = 1 (Y ≠ 0). So the answer is: X225 + Y29 = 1 (Y ≠ 0)
It is known that the vertex a coordinate of the triangle ABC is (2,1) point B moves on y = x, point C moves on X axis, and the minimum circumference of the triangle ABC is obtained
ABC is an isosceles right triangle (AC = BC, AC ⊥ BC),
That is, when B (1,1), C (3 / 2,0), the perimeter of the triangle is the smallest, and the minimum value is 1 + √ 2
In the known triangle ABC, point a (4,5), point B is on the x-axis, and point C is on the line L: 2x-y + 2 = 0. Find the minimum perimeter of the triangle ABC and the coordinates of a and B
Let D (x, y), then the point d (x, y) and the midpoint of point a (4,5) have X + 4-1 / 2 (y + 5) + 2 = 0 on the line 2x-y + 2 = 0, and the line ad must be perpendicular to the line 2x-y + 2 = 0, then (Y-5) / (x-4) = - 1 / 2
Given the point a (2,), find a point B on the x-axis, and find a point C on the line y = x to form a triangle ABC, so that its circumference is the smallest, and find the coordinates of points B and C
And the minimum perimeter
The small one has only 9 points, and got 5 points. I hope the experts don't dislike it,
Take the line y = x as the symmetry axis, make the symmetry point of point a, record D, its coordinate is (1,2). Take the X axis as the symmetry axis, make the symmetry point of point a, record e, its coordinate is (2, - 1). Triangle ABC circumference = AC + AB + BC = DC + CB + be ≥ DB + be ≥ De, de length
RELATED INFORMATIONS
- 1. There is a point a on the x-axis and a point B C (2,1) on the line y = X
- 2. In the triangle ABC, ab = 15, AC = 20, the height ad on the side of BC = 12, try to find the length of BC? I give bonus points to the detailed process, get 7 within the 25 do not talk about I want the process of 7
- 3. As shown in the figure, in △ ABC, be bisects ∠ ABC intersects AC at point E, de ‖ BC intersects AB at point D, ∠ ade = 70 ° and calculates the degree of ∠ DEB Don't use the formula (one outer angle of a triangle is equal to the sum of two nonadjacent outer angles),
- 4. In known quadrilateral ABCD, angle abd = 12 °, angle DBC = 36 °, angle ACB = 48 ° and angle ACD = 24 °, try to find the degree of angle ADB I can draw pictures according to the title,
- 5. It is known that in the oblique triangle ABC, the straight line with high BD and CE intersects at h, and the angle a = 45 degrees, so the degree of the angle BHC can be calculated? No picture, if you can,
- 6. As shown in Figure 7-11, in the known triangle ABC, ad is perpendicular to D, BC is equal to D, AE is equal to ∠ BAC (2), if ∠ B ＞ C, try to explain ∠ DAE = 1 / 2 (∠ B - ∠ C)
- 7. Ad is the height of triangle ABC, be is the midline angle on the side of AC, CBE = 30 degrees to prove ad = be
- 8. In rectangular ABCD, AE is perpendicular to BD, and the angle DAE = 2 angle BAE?
- 9. As shown in the figure, in RT △ ABC, ∠ C = 90 °, am is the middle line on the side of BC, sin ∠ cam = 35, then the value of tanb is () A. 32B. 23C. 56D. 43
- 10. As shown in the figure, in the isosceles trapezoid ABCD, it is known that ad ∥ BC, ab = CD, AE ⊥ BC are in E, ∠ B = 60 °, DAC = 45 ° and AC = 6?
- 11. As shown in the figure, in △ ABC, AE is the angular bisector, and ∠ B = 52 °, C = 78 ° to find the degree of ∠ AEB
- 12. It is known that, as shown in the figure, in the triangle ABC, ab = AC, point D is the midpoint of BC, De is perpendicular to AB and E, DF is perpendicular to AC and F, so it is proved that angle DFE = angle def
- 13. In the triangle ABC, the angle ACB = 90 degrees, AC = BC, M is a point in the triangle ABC, and am = 3, BM = 1, CM = 2
- 14. It is known that AB and CD intersect on O, m and N on AB, and AC = BD, am = BN, DM = CN
- 15. Draw the largest square in a right triangle. The known bottom is 10cm and the height is 30cm. Find the side length of the square Draw the largest square in a right triangle. The known bottom is 10cm and the height is 30cm. Find the side length of the square,
- 16. As shown in the figure, the middle angle ACB of triangle ABC is equal to 90 degrees, CA is equal to 15cm, CB is equal to 20cm, and the circle C with Ca as the radius intersects AB at d to find the length of AD
- 17. A right triangle, two of which are 6cm and 8cm respectively, is rotated along the hypotenuse 10cm for one circle to get a body of revolution?
- 18. If the waist length of an isosceles triangle with a vertex angle of 120 ° is 4cm, the diameter of its circumscribed circle is smaller______ cm．
- 19. The waist length of an isosceles triangle is 2, the vertex angle is 120 ° and the area of its circumcircle is calculated As the title
- 20. If the height on the bottom of an isosceles triangle is 3cm and the waist length is 3cm, the vertex angle is degree