It is known that the midpoint D.E of triangle ABC is a point on AB and AC respectively, and satisfies AD / DB = AE / EC = 1 / 2. How many times is BC equal to de? (not isosceles triangle) the process should be detailed

It is known that the midpoint D.E of triangle ABC is a point on AB and AC respectively, and satisfies AD / DB = AE / EC = 1 / 2. How many times is BC equal to de? (not isosceles triangle) the process should be detailed

Because it is known that D.E. at the midpoint of triangle ABC are AB.AC A point on the line, and we also know that AD / DB = AE / EC = 1 / 2, we can know AB = 1 / 2Ac, and we also know that this is an isosceles triangle, AC = BC. So ad = de. so BC = 4de

In △ ABC, ab = AC, the vertical bisector of AB intersects AB at point n, the extension of BC intersects at point m, ∠ a = 30 °
(1) Find the size of ∠ NMB
(2) If the degree of ∠ a in (1) is changed to 40 ° and other conditions remain unchanged, calculate the size of ∠ NMB
(3) What rules do you find? Try to prove them
(4) If ∠ a in (1) is changed to obtuse angle, should the understanding of this problem be modified?

① In △ ABC, ab = AC, the vertical bisector of AB intersects AB at n, the extension of BC intersects at m, and ∠ a = 40 ° to find the size of ∠ NMB
② For example, if the degree of ∠ a in (1) is changed to 70 ° and other conditions remain unchanged, then calculate the size of ∠ NMB
③ What rules do you find from the above conclusion? Try to prove it
④ If ∠ a in (1) is changed to obtuse angle, does this rule need to be modified

(1) ∵ - B = 1 / 2 (180 ° - a) = 75 °, ∵ - M = 15 °; (2) similarly, ∵ - M = 35 °; (3) the rule is: the size of ∵ - M is half of the size of ∵ - A, that is, the acute angle between the vertical bisector of AB and the bottom BC is equal to half of ∵ - A

In the triangle ABC, ad is the bisector of the angle BAC, and E is the point on the extension of AD, if the square of EC = ed * EA
Verification: ab * AC = ad * AE
There should be a complete process

Certification:
EC*EC=ED*EA
Angle e = angle e
EDC is similar to ECA
=>
Angle ECB = angle CAE = angle BAE
AC/CD=EC/DE…… one
=>
Bad is similar to ECD
=>
AB/CE=AD/CD…… two
1 * 2
AC*AB = CE*CE*AD/DE
Known
CE*CE=DE*AE
=>
AC*AB=DE*AE*AD/DE
=AD*AE

In the triangle ABC, a + B = 2c, angle a = angle c + 60 ° to find SINB

A + B = 2C. From the sine theorem, we get Sina + SINB = 2sinc. We know that a = B + 60, C = 120-2b. We can get sin (B + 60) + SINB = 2Sin (120-2b) from the formula

In the triangle ABC, if a + B = (greater than or equal to) 2c, it is proved that C = (less than or equal to) 60 degrees

Using cosine formula directly
cosC=(a^2+b^2-c^2)/(2ab)≥[a^2+b^2-((a+b)/2)^2]/(2ab)
=(3a^2+3b^2-2ab)/8ab
≥(6ab-2ab)/8ab=1/2
And C ∈ (0180 °), so
C≤60°

In triangle ABC, if angle c = 90 degrees, Tana = 3, then SINB=

tanA=3,
tanA=sinA/cosB=BC/AC=3.
cosA=3sinA,
And sin ^ 2A + cos ^ 2A = 1,
sinA=3√10/10,cosA=√10/10,
That is: AC = 3 √ 10, BC = √ 10, ab = √ (AC ^ 2 + BC ^ 2) = 10
sinB=AC/AB=3√10/10.

In △ ABC, ∠ C = 90 °, Tana = 1 / 3, find the value of cosa, SINB

Because ∠ C = 90 °, so △ ABC is RT triangle,
And Tana = 1 / 3, so a / b = 1 / 3,
According to Pythagorean theorem, C / a = √ (1 ^ 2 + 3 ^ 2) = √ 10,
And cosa = B / C = 3 / √ 10 = 3 √ 10 / 10 = SINB,
So cosa = SINB = 3 √ 10 / 10

In the triangle ABC, the angle c is 90 ° Tana is one-third of SINB

tanA=BC/AC=1/3
Let BC = k, then AC = 3K
According to Pythagorean theorem, ab = radical 10 * k
So SINB = AC / AB = 3K / (radical 10 * k) = 3 / radical 10 = 3 times radical 10 / 10

In △ ABC, ∠ C = 90 ° and Tana = 1, then the value of SINB is ()
A. 3B. 2C. 1D. 22

In ∵ △ ABC, ∵ C = 90 °, Tana = 1, ∵ a = 45 °, ∵ B = 180 ° - ∵ a - ∵ C = 45 °, ∵ SINB = sin45 ° = 22