Three digit ABC, the sum of all the two digits of a, B and C, multiplied by 1 / 3, is the three digit ABC =?

Three digit ABC, the sum of all the two digits of a, B and C, multiplied by 1 / 3, is the three digit ABC =?

22a+22b+22c=300a+30b+3c
278a+8b=19c
19c

a. B, C are natural numbers, abc-ab-c = 195, ask ABC?

195+1=a ,195-1=b,195=c

Find a mathematical problem: let △ ABC three sides a, B, C length are natural numbers, and a ≤ B ≤ C, a + B + C = 13, how many triangles with a, B, C as sides

2c＜a＋b＋c＝13
∴c＜6.5
That is, the maximum value of C is 6
And 13 = a + B + C < 3C
C > 13 / 3
The minimum value of C is 5
When C = 6, B = 6, a = 1
When C = 5, B = 5, a = 3 or B = 4, a = 4
Therefore, there are three triangles satisfying the above conditions

If ABC is any three integers, then there are () integers of 1 / 2 × (a + b), 1 / 2 × (c + b), 1 / 2 (a + C)

3

Judge the position relation of the following points and give the proof (1). A (1,2) B (- 3, - 4) C (2,3.5)
Judge the position relation of the following points and give the proof
(1).A(1,2)B(-3,-4)C(2,3.5)
(2)P(-1,2)Q(0.5,0)R(5,-6)

Try using space rectangular coordinate system

Given a (0,1), B (1,0), C (1,2), D (2,1), judge the position relationship between AB and CD, and give the explanation
In learning vector, collinear and so on for explanation!

Drawing is the fastest. Positive solution on the trouble point, using vector
Vector AB = B-A = (1, - 1)
Vector CD = d-c = (1, - 1)
So AB and CD are parallel (when vector a = k * vector B, the two vectors are parallel or collinear, K is constant.) here k = 1

Given the points a (0,1), B (1,0), C (1,2), D (2,1), try to judge the positional relationship between the vectors AB and CD, and give the proof

AB and CD are collinear. Prove: because AB = (1, - 1), CD = (1, - 1), so AB = CD. So AB and CD are collinear

Known: as shown in the figure, a, B, C three points on the same line, angle 1 = angle 2, angle d = angle 3, prove BD parallel CE

You don't have letters on the picture
Let me assume that the following three are ABC
Corner two and three, there's E
Because angle 1 equals angle 2
So ad is parallel to be
So angle D equals angle DBE
And because angle D is equal to angle 3
So the angle DBE is equal to the angle 3
So BD is parallel to CE
It's a zero point question
I can't. I have to add some points

It is known that Ca = CB = CD, the circle AB passing through a, C and D is at point F. it is proved that CF is the bisector of ∠ DCB

It is proved that: connecting DF, BD, ∵ AC = CB = CD, ∵ a = ∵ 2, ∵ CDB = ∵ CBD, and ∵ a = ∵ 1, ∵ 1 = ≌ 2, ∵ FDB = ∵ FBD, ? DF = BF, in △ DCF and △ BCF, ? DF = BF, ? 1 = ≌ 2, CD = CB, ≌ DCF ≌ BCF, ≌ DCF = ≌ BCF, CF is the bisector of ∠ DCB

As shown in the figure, AC = BD, ∠ a = ∠ B, points E and F are on AB, and de ‖ CF, try to explain that this is a centrosymmetric figure

Connecting CD, AB to o. ∵ in △ ACO and △ BDO, ∠ COA ≌ DOB ≌ a ≌ BAC = BD, ≌ ACO ≌ BDO (AAS), so OA = ob, OC = OD. ≌ de ∥ CF, ≌ deo = ∥ CFO, in △ ode and △ OCF, ≌ Deo ≌ CFO ≌ doe ≌ Cofod = OC, ≌ ode ≌ OCF (AAS)