2013 China Normal University self recruitment question 2 ·· proving A2 + B2 + C2 + D2 = 9r2 I feel ashamed to die in the second question··

The title is wrong & nbsp; it should be from the center to the outside
The first reaction of this problem is to calculate the relationship between D and triangle, and then substitute it into A2 + B2 + C2 + D2 test. However, it will be found that D is not good-looking (because to determine d, I use three points of the center of gravity perpendicular to the epicenter to be collinear and the ratio is 1:2, and the distance between the center of gravity perpendicular to the epicenter is directly opposite to an edge to make a perpendicular parallel line, and then Pythagorean theorem)
Another idea is vectors,
ABC center O center O center O center O center O center O center h, then Oh = OA + ob + OC (all these are vectors & nbsp; the same below) (this point can be directly calculated (oh-oa) * (ob-oc) = 0 verification, in other words, you can first make such a h, and then he is a center) & nbsp; then AB & \35\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\3 (OA & # 178; + ob & # 178; + OC & # 178;) = 9R & # 178;

G is the center of gravity of triangle ABC. Try to explain s triangle BDG = s triangle BFG = s triangle APG

∵ G is the center of gravity of △ ABC
The central lines of △ ABC are ad, be and CF
∴AF=BF,BD=CD,AE=CE
G as GH ⊥ AB over H
∵S△AFG=1/2*AF*GH,S△BFG=1/2*BF*GH
∴S△AFG=S△BFG
Also ∵ AE = CE ∵ s △ Abe = s △ CBE, s △ AEG = s △ CEG (equal base and equal height)
∴S△ABG=S△ABE-S△AEG=S△CBE-S△CEG=S△CBG
∵S△AFG=1/2*S△ABG,S△BDG=1/2*S△CBG
∴S△AFG=S△BDG
∴S△BDG=S△BFG=S△AFG
Absolutely original. Feel good to add points Oh

In the triangle ABC, the opposite sides of angles a, B and C are a, B and C respectively, and B & # 178; = AC, then the value range of B is_______ .

Give the answer now!

Finding the size of a in triangle ABC by B ^ 2 + C ^ 2-A ^ 2 = BC
If Sina ^ 2 + SINB ^ 2 = sinc ^ 2, find the size of B

In the triangle ABC, B & # 178; + C & # 178; - A & # 178; = BC, find the size of a; if Sin & # 178; a + Sin & # 178; b = Sin & # 178; C, find the size of B
The solution is as follows: (1). Cosa = (B & # 178; + C & # 178; - A & # 178;) / (2BC) = BC / (2BC) = 1 / 2, so a = 60 & # 186;;
(2).sin²B=sin²C-sin²A=(sinC+sinA)(sinC-sinA)
=2sin[(C+A)/2]cos[(C-A)/2]×2cos[(C+A)/2]sin[(C-A)/2]
=2sin[(180º-B)/2]cos[(C-A)/2]×2cos[(180º-B)/2]sin[(C-A)/2]
=2cos(B/2)cos[(C-A)/2]×2sin(B/2)sin[(C-A)/2]
=sinBsin(C-A)
So SINB [SINB sin (C-A)] = 0
∵ SINB ≠ 0, there must be SINB sin (C-A) = 0, that is, SINB = sin (C-A); so B = c-a
That is, a + B = 180 & # 186; - C = C, 2C = 180 & # 186;, C = 90 & # 186;, a + B = 90 & # 186;, B = 90 & # 186; - A;
[the condition is not enough, the size of B cannot be determined]

In the triangle ABC, the angles ABC faces are ABC, satisfying B ^ 2 = AC, and a ^ 2-C ^ 2 = AC BC. Find the size of ∠ a
The value of bsinb / C

B ^ 2 + C ^ 2-A ^ 2 = BC
And cosa = (b ^ 2 + C ^ 2-A ^ 2) / 2BC = 1 / 2
So in a triangle, a = π / 3
From B ^ 2 = AC, we can get (SINB) ^ 2 = sinasinc, B / C = SINB / sinc
bsinB/c=(sinB)^2/sinC=sinA=√3/2

In triangle ABC, if (a + B + C) (B + C-A) = BC, what is the size of angle a?

（a+b+c)(b+C-A)=bc,
That is, (B + C) ^ 2-A ^ 2 = BC
Then B ^ 2 + C ^ 2-A ^ 2 = - BC
cosA=(b^2+c^2-a^2)/2bc=-1/2
Then a = 120 degrees

In the triangle ABC, a, B and C are the opposite sides of angles a, B and C respectively, and satisfy B ^ 2 + C ^ 2 = a ^ 2 + BC. There is a detailed process to find the size of angle A
Such as the title

This question is about the application of cosine theorem. The specific formula is as follows: (if you don't learn cosine theorem, write it down)
In triangle ABC, if a, B and C are opposite sides of angles a, B and C respectively, then a & sup2; = B & sup2; + C & sup2; - 2bccosa, B & sup2; = A & sup2; + C & sup2; - 2accosb,
c²=a²+b²-2abcosC
So from the above formula A & sup2; = B & sup2; + C & sup2; - 2bccosa: cosa = (B & sup2; + C & sup2; - A & sup2;) / 2BC = BC / 2BC = 1 / 2
So the angle a is 60 degrees

It is known that a, B and C are the trilateral lengths of △ ABC respectively. Try to compare the sizes of (a * 2 + b * 2-C * 2) * 2 and 4A * 2B * 2

It is known that a, B and C are the lengths of the three sides of △ ABC

∵ a2-b2-c2-2bc = a2 - (B + C) 2 = (a-b-c) (a + B + C) ∵ a, B, C are three sides of triangle, ∵ a − B − C < 0, ∵ A2 − B2 − C2 < 2BC

In △ ABC, it is known that ∠ A: B: C = 2:3:7, and the degree of each angle is calculated

The sum of the internal angles is 180 degrees
Therefore, a = 180 × 2 (2 + 3 + 7) = 30 degrees
∠ B = 30 ÷ 2 × 3 = 45 degrees
∠ C = 180-30 + 45 = 105 degrees

2013 China Normal University self recruitment question 2 ·· proving A2 + B2 + C2 + D2 = 9r2 I feel ashamed to die in the second question··