As shown in the figure, △ ABC, ab = AC, points D, e and F are on edges BC, AB and AC respectively, and BD = CF, ∠ EDF = ∠ B. is there a triangle congruent with △ BDE in the figure? And explain the reason

As shown in the figure, △ ABC, ab = AC, points D, e and F are on edges BC, AB and AC respectively, and BD = CF, ∠ EDF = ∠ B. is there a triangle congruent with △ BDE in the figure? And explain the reason

There are ≌ BDE ≌ △ CFD. Reasons: ∤ EDC = ≌ EDF + ≌ CDF, ∤ EDC = ∤ B + ∤ bed, ∤ EDF + ≌ CDF = ≌ B + ≌ bed, and ∤ EDF = ∤ B, ∤ bed = ∤ CDF. ∤ AB = AC ∤ B = ∤ C ∤ BD = CF ≌ BDE ≌ CFD (AAS)

It is known that △ ABC is an equilateral triangle, BD is the middle line, extend BC to e, make CE = CD = 1, connect De, then De=______ ．

In RT △ BDC, BD = 22 − 12 = 3, i.e. de = B, BD = 22 − 12 = 3, i.e. in RT △ BDC, BD = 22 − 12 = 3, i.e. de = B. in RT △ BDC, BD = 22 − 12 = 3, i.e. in RT △ BDC, BD = 22 − 12 = 3, i.e. in RT △ BDC, BD = 22 − 12 = 3, i.e. in RT △ BDC, BD = 22 − 12 = 3, i.e. in RT △ BDC, BD = 22 − 12 = 3, i.e. de = B D = 3, so the answer is: 3

Given cot a = m, find the value of sin a and COS a

cos a/ sin a=cot a=m
cos a=m sin a
cos²a+sin²a=1
(1+m²)sin²a=1
sina=√[1/(1+m²)]
cosa=m√[1/(1+m²)]

Given sin (π + α) = 1 / 4, the value of sin (2 π - α) - cot (α - π) cos α is obtained

sin(π+α)=1/4,sina=-1/4sin(2π-α)-cot(α-π)cosα=-sina+cot(∏-a)cosa=-sina-cotacosa=-sina-cosa/sina*cosa=1/4-(cosa)^2/(-1/4)=1/4+4[1-(sina)^2]=1/4+4(1-1/16)=4

Simplification: cos (a - π / 2) · cot (- A-3 π / 2) · color (- A + 5 π / 2) · Tan (5 π / 2 + a),

Is that "color" sec Because: cot = 1 / Tan sec = 1 / cos, the above formula can be written as: cos (a - π / 2) · sec (- A + 5 π / 2) · cot (- A-3 π / 2) · Tan (5 π / 2 + a) = cos (a - π / 2) · sec (- A + π / 2) · Tan (π / 2 + a) · cot (- A-3 π / 2) = C

It is known that in △ ABC, the lengths of edges corresponding to angles a, B and C are a, B and C respectively, the radius of circumcircle is 1, and the condition 2 (sin2a-sin2c) = (Sina SINB) B is satisfied, then the maximum area of △ ABC is______ ．

From the sine theorem, we can get b = 2rsinb = 2sinb. Substituting the known equation, we can get 2sin2a-2sin2c = 2sinasinb-2sin2b, sin2a + sin2b-sin2c = sinasinb, ∵ A2 + b2-c2 = AB, ∵ COSC = A2 + B2 − c22ab = 12, ∵ C = 60 °.∵ AB = A2 + b2-c2 = A2 + B2 - (2rsinc) 2 = A2 + b2-3 ≥ 2ab-3

In the triangle ABC, if B ^ 2Sin ^ 2C + C ^ 2Sin ^ 2B = 2BC · cos · bcos · C, try to judge the shape of the triangle

According to the sine theorem, the original formula can be reduced to sin ^ 2bsin ^ 2C + sin ^ 2csin ^ 2B = 2sinbsinccosbcosc
2sin^2Csin^2B=2sinBsinCcosBcosC
sinBsinC=cosBcosC
cosBcosC-sinBsinC=0
cos(B+C)=0
B + C = 90 degrees, so a = 90 degrees
So it's a right triangle

2Sin ^ 2B / 2 + 2Sin ^ 2C / 2 = 1, judge the shape of triangle

2sin^2B/2+2sin^2C/2=1
1-sinB+1-sinC=1
sinB+sinC=1
This condition is not enough to judge

In the triangle ABC, B ^ 2 = 4A ^ 2Sin ^ 2B, find a

∵b^2=4a^2sin^2B
∴b=2asinB
b/sinB=a/(1/2)
From the sine theorem B / SINB = A / Sina
∴a/(1/2)=a/sinA
∴sinA=1/2
A = π / 6 (or 30 & # 186;)

In the triangle ABC, the opposite sides of angles a, B and C are a, B and C respectively, and a (1 + COSC) + C (1 + COSA) = 3B
(1) To prove that a, B and C are equal difference sequence
(2) Finding the minimum value of CoSb