In the right triangle ABC, the angle ACB = 90 °, AC = BC, D is the midpoint of BC, CE is perpendicular to ad, BF is parallel to AC Verification of AB vertical bisection DF Point F is on the extension line of CE and BF

In the right triangle ABC, the angle ACB = 90 °, AC = BC, D is the midpoint of BC, CE is perpendicular to ad, BF is parallel to AC Verification of AB vertical bisection DF Point F is on the extension line of CE and BF

(the intersection of AB and DF is j) because D is the midpoint of BC, ad is the bisector of angle cab, so angle CAD = 22.5 degrees, because angle CAD = 22.5 degrees, angle acd90 degrees, so angle CDE = 67.5 degrees, because CE is perpendicular to ad, angle CDE = 67.5 degrees, so angle ECD = 22.5 degrees

⊿ ABC is an isosceles right triangle, ∠ C = 90 degrees, D is a point on BC extension line, CD = CE, e point is on AC, the extension line of be intersects ad with F. it is proved that BF is perpendicular to AD

Connect de and extend, intersect AB at G
Since ⊿ ABC is an isosceles right triangle, ⊿ C = 90 degrees
So ∠ cab = 45 degrees
Because ⊿ ABC is an isosceles right triangle, so ⊿ ACD = 90 degrees, plus CD = CE, so ⊿ CDE = 45 degrees
So ∠ cab = ∠ CDE
Because the antiparietal angle ∠ AEG is equal to ∠ Dec,
So ⊿ AEG is similar to ⊿ Dec
∠ age = ∠ DCE = 90 degrees
DG is perpendicular to AB, that is, De is perpendicular to ab
Because AC is perpendicular to DB, that is, AE is perpendicular to DB
So e is the vertical center of ⊿ ADB,
So be is perpendicular to AD
BF is perpendicular to AD

It is known that in the right triangle ABC, the angle ACB = 90 °, AC = 2, BC = 1, point D is on AB, CD = CB. If point E is on the extension line of CB and the triangle composed of three points a, B, e is similar to the triangle ACD, the length of be is calculated

Analysis: there should be two cases of similarity, one is for BAE = ∠ ACD, the other is for BAE = ∠ CAD,
∵CD=CB,∴∠CDB=∠CBD,∴∠ADC=∠ABE,
1) When BAE = ACD, △ ACD ∷ EAB,
/ / AC / AE = CD / AB, i.e. 2 / √ [4 + (1 + x) ^ 2] = 1 / √ 5,
The solution is x = 3
2) When BAE = CAD, △ ACD ∽ AEB
/ / AC / AE = CD / be, i.e. 2 / √ [4 + (1 + x) ^ 2] = 1 / X
When 3x ^ 2-2x-5 = 0, x = 5 is obtained, and the other negative value is discarded
In conclusion, be = 3 or 5

In △ ABC, if the angles a, B and C form an arithmetic sequence, and B = 2, then the radius r of the circumscribed circle is r=______ .

∵ in ᙽ ABC, the angles a, B, C form an arithmetic sequence,
∴2B=A+C，
∵A+B+C=180°，
∴B=60°，
∵b=2，
From the sine theorem B
SINB = 2R, r = B
2sinB=2
2 x
Three
2=2
Three
3．
So the answer is: 2
Three
Three

The circumscribed circle radius of triangle ABC is r, and the angle c is 60 degrees. The value range of (a + b) / R?

According to the sine theorem: a = 2rsina; b = 2rsinb;
So (a + b) / r = 2 (Sina + SINB) = 2 × 2Sin [(a + b) / 2] cos [(a-b) / 2] = 4sin ((π - 60 °) / 2) cos ((a-b) / 2) = 2 √ 3cos ((a-b) / 2)
Because of 0

The circumcircle radius of triangle ABC is r = 2, a: B = 3:4, C = 60 degrees Then a=_____ ,b=______ .

Let a / b = 3 / 4 = k then, a = 3k, B = 4K, B = 4K, B = 4K by the cosine theorem: C ^ 2 = a ^ 2 + B ^ 2-2abconb that is, (3K ^ 2 + (4K)) ^ 2-2 * 3K * 4kcon60 = (root number 3) ^ 29K ^ 2 + 16K ^ 2-12k ^ 2 = 313k ^ 2 = 3K ^ 2 = 3 / 13K = (root number 39) / 13, therefore, a = 3 (root number 39) / 13b = 4 (root number 39) / 13b = 4 (root number 39) / 13b = 4 (root number 39) / 13b = 4 (root number 39) / 13b = 4 (root number 39) / 13b = 4 (root number 39) / 13b = 4 (root number 39) / 13b = 4 (root number 39

Given that the side length of the triangle ABC is A.B.C, the area is s, ABC = 1, s = 2, find the radius of the circumscribed circle of the triangle

S=1/2absinC
Sine theorem a / Sina = B / SINB = C / sinc = 2R
So s = ABC / 4R
R=abc/4S=1/8

In the triangle ABC, the angle a = 45 °, the angle B: the angle c = 4:5, and the longest side length is 10

The circumscribed circle has a radius of 10 √ 3 / 3 and an area of 100 π / 3. Firstly, the degrees of angles B and C are 60 ° 75 ° respectively. AB side is 10, and triangle is acute angle. The center of circumscribed circle is determined as o point. The connection of OA ob OC angle BOC is 2 times angle. A is 90 ° Sina / SINB = A / b a = BC = √ 6 / 3B ob = √ 3 / 3B sin angle Bao /

In △ ABC, the edges to which the inner angles a, B and C are opposite are a, B, C respectively, and the circumcircle radius is 6, B 1−cosB=24，sinA+sinC=4 3． (1) Find CoSb; (2) Find the maximum area of △ ABC

(1) B1 − CoSb = 24 ‰ 2 × 6sinb1 − CoSb = 24 ν 2 (1-cosb) = SINB (3 points)  4 (1-cosb) 2 = sin2b = (1-cosb) (1 + CoSb) ∵ 1-cosb ≠ 0,  4 (1-cosb) = 1 + CoSb,  CoSb = 35, (6 points) (2) ? Sina + sinc = 43, ? A12 + C12 = 43

In the triangle ABC, the opposite sides of the three inner angles a, B and C are respectively a, B, C, and the circumcircle radius of the triangle ABC is known to be r If 2R (sin2a-sin2c) = (root 2a-b) SINB (1) find the size of angle c (2) find the maximum area of the triangle ABC

Because a = 2rsina B = 2rsinb C = 2rsinc
It is known that 2R (sin? - sin?) = (√ 2a-b) SINB
Multiply the two sides by 2R to obtain a ^ + B ^ - C 2 = √ 2Ab
So COSC = √ 2 / 2 C = 45 degrees
S=(1/2)absinC=√2R²sinAsinB
=(√2/2)cos(A-B)+1/2
ν Smax = (√ 2 + 1) / 2