Answers to Tianfu mathematics Issue 8, 2010 In the triangle ABC, the angle BAC = 90 degrees, ad is the height on the BC side, e is a moving point on the BC side (not coincident with points B and C), EF is perpendicular to BC, eg is perpendicular to AC, and the perpendicular foot is f and g respectively (1) Verify eg / ad = CG / CD; (2) When connecting FD and DG, please judge whether FD and DG are vertical? If so, please give proof; (3) Is the triangle FDG isosceles right triangle when AB = AC? Why?

Answers to Tianfu mathematics Issue 8, 2010 In the triangle ABC, the angle BAC = 90 degrees, ad is the height on the BC side, e is a moving point on the BC side (not coincident with points B and C), EF is perpendicular to BC, eg is perpendicular to AC, and the perpendicular foot is f and g respectively (1) Verify eg / ad = CG / CD; (2) When connecting FD and DG, please judge whether FD and DG are vertical? If so, please give proof; (3) Is the triangle FDG isosceles right triangle when AB = AC? Why?

If EF is perpendicular to BC, eg is perpendicular to AC, and the foot drop is f and g respectively. Since EF is perpendicular to BC, how can the foot drop be f? The answer to the first question is to prove that the proof of EG / ad = CG / CD is equivalent to the proof of eg / CG = ad / CD

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One side of the rectangle is 3M + 2n, and the other side is M-N longer than it

(3m + 2n) (3m + 2n + m-n) = 12m2 + 11mn + 2n2. A: the area of a rectangle is 12m2 + 11mn + 2n2

Let g and m be the center of gravity and outer center of triangle ABC, a (- 1,0), B (1,0), and vector GM and vector AB are parallel, the trajectory of C is e, the two upper and lower intersections of E and Y axis are A2, A1, the moving points m and n are all on e, and satisfy vector a1m point multiplied by vector A1N = 0, whether the intersection point P of line A1N and A2M is constant on a certain line L, if not, try to find the equation of L; if not, please explain the reason

Let C (x, y), G (x / 3, Y / 3), then M (0, Y / 3)
Cm = am
So x ^ 2 + (2 / 3Y) ^ 2 = 1 ^ 2 + (Y / 3) ^ 2
That is e: x ^ 2 + y ^ 2 / 3 = 1
Let m (x0, Y0), the slope of A2M, a1m and A1N be K1, K2 and K3 respectively
Deformation from x0 ^ 2 + Y0 ^ 2 / 3 = 1: (*)
[(y0-√3)/x0]*[(y0+√3)/x0]=-3
That is, K1 * K2 = - 3
And K2 * K3 = - 1
So we can make K1 = 3K3 = K
Let A2M: y = KX + √ 3
Let A1N: y = (K / 3) x - √ 3
By combining A2M and A1N, P (- 3 √ 3 / K, - 2 √ 3) is obtained
That is, P is constant on the line y = - 2 √ 3
PS: (*) formula described in the deformation can be extended to any standard equation of ellipse and hyperbola, LZ can try~~~~

Given that the coordinates of vertex A and B of △ ABC are a (0.0) B (6.0) and vertex C moves on the curve y = x ^ 2 + 3, the trajectory equation of the center of gravity of △ ABC is obtained

Let C (a, a ^ 2 + 3)
Then the coordinates of the center of gravity are (x, y)
There are: x = (0 + 6 + a) / 3 = 2 + A / 3
y=(0+0+a^2+3)/3=1+a^2/3
Therefore: A ^ 2 = 9 (X-2) ^ 2 = 3 (Y-1)
That is y = 3 (X-2) ^ 2 + 1

It is known that the vertex of triangle ABC is a (0,1) B (8,0) C (4,10) if the vector BD = vector DC, and the vector CE = 2, and the vector EA, ad and be intersect with F,
Find vector AF

Because vector BD = vector DC, so D is the midpoint of BC (6,5), vector CE = 2, vector EA, then E is the trisection point (4 / 3,4) of AC, and then through the two-point formula (Y-Y1) / (y2-y1) = (x-x1) / (x2-x1), we can respectively calculate the linear equations of AD and BC, and simultaneously solve the equations to find the point F coordinate. Then we can get the vector AF

In △ ABC, the vector BD = vector DC, the vector CE = 2, the vector EA, and the line segment AD and be intersect at F
Let AB = a, AC = B, ad = P, be = Q;
(1) Let a and B denote P and Q respectively; (2) let P and Q denote a; (3) if BF = λ Q, find the value of λ

(1) Vector BD = DC
P = vector ad = (1 / 2) AB + (1 / 2) AC = (a + b) / 2
Vector CE = 2EA,
Ψ q = vector be = ae-ab = (1 / 3) ac-ab = - A + B / 3. ②
(2)①*2-②*3,2p-3q=4a,
∴a=p/2-(3/4)q.
(3) If DG ∥ be intersects AC with G, then CG = Ge = EA, DG = be / 2, Fe = dg / 2 = be / 4,
∴BF=3BE/4=(3/4)q,
∴λ=3/4.

If the vertices a (1, - 1,2), B (5, - 6,2), C (1,3, - 1) of △ ABC, then the length of high BD on the edge of AC is equal to?
From points a (1, - 1,2) and C (1,3, - 1), it can be assumed that the coordinates of point D are (x, y, z), where
x＝1*t＋1*(1－t)＝1,
y＝-1*t＋3*(1－t)＝3－4t,
z＝2*t＋(－1)*(1－t)＝3t－1.
Why (1-T)?

Since point D is on a straight line AC, we can use a parameter t to represent the coordinates of point D. from points a (1, - 1,2) and C (1,3, - 1), we can set the coordinates of point D as (x, y, z), where x = 1 * t + 1 * (1-T) = 1, y = - 1 * t + 3 * (1-T) = 3-4t, z = 2 * t + (- 1) * (1-T) = 3t-1

If we know the vertex of triangle ABC, a (1, - 1,2) B (5, - 6,2) C (1,3, - 1), then what is the high BD on the edge of AC?

AB = {(5-1) ^ 2 + [- 6 - (- 1)] ^ 2 + (2-2) ^ 2} = √ 41. AC = {[(1-1) ^ 2 + [3 - (- 1)] ^ 2 + (- 1-2) ^ 2]} = 5. BC = {(1-5) ^ 2 + [3 - (- 6)] ^ 2 + (- 1-2) ^ 2} = √ 106. In △ ABC, cosine theorem is applied: cosa = (AC ^ 2 + AB ^ 2-bc ^ 2) / 2Ab* AC.cosA= (25+41-106)(/2*5* √41...

Given the vertices a (1, - 1), B (5, - 6), C (1,3) of △ ABC, we can find the high BD on the edge of AC

The equation of AC: x = 1, parallel to the Y axis, BD should be parallel to the X axis, passing through point B, so the equation is y = - 6, | BD | = | 5-1 | = 4