In the acute triangle ABC, AB > AC, CD and be are the heights on the sides of AB and AC respectively. The extension line of De and BC is called t, and the vertical line of BC crossed by D, BC crossed by F, It is proved that f, G and T are collinear

In the acute triangle ABC, AB > AC, CD and be are the heights on the sides of AB and AC respectively. The extension line of De and BC is called t, and the vertical line of BC crossed by D, BC crossed by F, It is proved that f, G and T are collinear

From the meaning of the title: F, G, t are not collinear
Because it is clear in the title that point F and T are on the straight line BC, and point G is on the high Cd on the edge of ab,
Therefore, it is impossible to prove that f, G and T are collinear

In the acute triangle ABC, the known angles a = α, CD be are the high proof de = bccos α on AB, AC respectively

∵∠A=∠A,∠ADC=∠AEB=90°
∴△ADC∽△AEB
∴AD:AE=AC:AB
That is ad: AC = AE: ab
∵∠A=∠A
∴△AED∽△ABC
∴DE:BC=AE:AB=cosα
∴DE=BCcosα

It is known that in the acute triangle ABC, ab = AC, BC = 4, D is a point on the edge of AC, ad: DC = 3:1, Sina = 12 / 13,
Find the length of CD and the area of triangle BCD

(1) AB = AC, we can see that the triangle is isosceles triangle, through a to BC vertical line, perpendicular foot is e, then the angle EAC = 1 / 2 angle A
Because Sina = 12 / 13, we can see that cosa = 5 / 13. According to the half angle formula: cosa = 1-2sin (A / 2) ^ 2 = 5 / 13
The solution is sin (A / 2) = sin angle EAC = EC / AC = 2 √ 13 / 13, known EC = 1 / 2BC (isosceles triangle property) = 2
It can be seen that AC = √ 13 is due to AD: DC = 3:1, that is, CD = 1 / 4ac = √ 13 / 4
(2) In triangle BCD and triangle ABC,
The height is the same as the distance from B to AC, the bottom edge is 1:3, CD: ad = 1:3
So the area ratio is 1:3
S triangle ABC = 1 / 2 * AB * ac * Sina, AC = 4DC
It can be obtained by substituting the data
Area of triangle BCD = 2

As shown in the figure, it is known that in isosceles △ ABC, ab = AC = 20cm, the perpendicular line of AB intersects the other waist at point D, and the circumference of △ BCD is 30cm, then BC=____

The perpendicular of AB intersects the other waist at point D
∴AD＝BD
The circumference of ∧ BCD = BD + CD + BC = AD + CD + BC = AC + BC
The circumference of ∵ △ BCD = 30, AC = 20
∴20+BC＝30
∴BC＝10（cm）

As shown in the figure, in ABC, ab = AC = 20cm, BC = 15cm, the vertical bisector of Ba intersects AC at point D, intersects AB at point E, and calculates the perimeter of BCD

De is the vertical bisector of AB, so ad = dB
The circumference of △ BCD = BC + DC + BD
BC=15 DB+DC=AD+DC=AC
So the circumference of △ BCD = 35

sin²A+sin²B=cos²C

sin²A+sin²B=cos²C
(1-cos2A)/2+(1-cos2B)/2=1-sin^2C
cos2A+cos2B=2sin^2C
2cos(A+B)*cos(A-B)=2sin^2C
-sinC*cos(A-B)=sin^2C
cos(A-B)=-sinC sinC>0
cos(A-B)π/2
So one of a or B is obtuse
It's an obtuse triangle

There are 11 obtuse angles, 15 right angles and 100 acute angles in a triangle. How many acute angles are there?
After a math group activity, Xiao Ming cleaned up the triangle model on his desk. After checking, there are 11 obtuse angles, 15 right angles and 100 acute angles. How many acute angles are there?

Because there is at most one obtuse angle or right angle in a triangle, then 11 obtuse angles are equivalent to 11 obtuse angle triangles, then there are 2 acute angles in each obtuse angle triangle; if there are 15 right angles, then there are 2 acute angles in each right angle triangle; now there are 15 + 11 = 26 triangles

In triangle ABC, if Tan β = cos (C-B) / Sina + sin (C-B), judge the shape of triangle?
Please give the inference process!

Sinbsina + sinbsin (C-B) = cosbcos (C-B)
Sinbsina + sincsinbcosb COSC (SINB) ^ 2 = (CoSb) ^ 2 * COSC + sincsinbcosb
Both sides merge COSC ((SINB) ^ 2 + (CoSb) ^ 2) = sinasinb
That is, COSC = sinasinb
And because COSC = - cos (a + b) sinasinb = - cos (a + b)
Cosacosb = 0
So in a triangle, a = 90 or B = 90, that is, ABC is a right triangle
Method 2
tanB=cos(C－B)/〔sinA+sin(C－B)〕＝cos(C－B)/〔sin（B+C)+sin(C－B)〕
tanB=(cosBcosC+sinBsinC)/(2sinCcosB)
2sinBsinC=cosBcosC+sinBsinC
cosBcosC-sinBsinC=0
cos(B+C)=0
cosA=0
A = 90 degrees
That is, ABC is a right triangle
Reference: Baidu

In the triangle ABC, the opposite sides of the angles a, B and C are a, B, C and Tan respectively. The angle c is equal to 63 under the root sign. (1) find the cos angle C
(2) If the vector CB times the vector CA equals 5 / 2, and a + B is 9, find C

tgc=sinc/cosc
=Root sign (square of 1-cosc) / COSC
63 under root = under root (square of 1-cosc) / COSC
COSC=1/8
More than two

In the triangle ABC, a, B and C are the opposite sides of angles a and BC respectively, cos a = √ 5 / 5, Tan B = 3

Cosa = radical (5) / 5
Sina = 2 radical (5) / 5 (Sina) ^ 2 = 1 - (COSA) ^ 2
tanB=3
SINB = 3 / radical (10) (1 / cos b) ^ 2 = 1 + (tanb) ^ 2
CoSb = 1 / radical (10)
cosC=cos(pi-A-B)=-cos(A+B)
=-(cos A*cos B-sin A*sin B）
=-(1 radical (5) / 5 * 1 / radical (10) - 2 radical (5) / 5 * 3 / radical (10))
=1 / radical (2)
C=pi/4