(1 / 3) the known triangles ABC, (|) BD and CE are bisectors of angle B and angle c, respectively. Through point a, make AF vertical BD, Ag vertical CE, the perpendicular feet are f, G, connecting FG (1 / 3) the known triangles ABC, (|) BD and CE are bisectors of angle B and angle c, respectively. Through point a, make AF vertical BD, Ag vertical CE, and the perpendicular feet are f, G, connecting FG. (2) if C

(1 / 3) the known triangles ABC, (|) BD and CE are bisectors of angle B and angle c, respectively. Through point a, make AF vertical BD, Ag vertical CE, the perpendicular feet are f, G, connecting FG (1 / 3) the known triangles ABC, (|) BD and CE are bisectors of angle B and angle c, respectively. Through point a, make AF vertical BD, Ag vertical CE, and the perpendicular feet are f, G, connecting FG. (2) if C

Answer tip: let BC be the largest edge
1. Let the extension lines of Ag and AF intersect BC at M and N respectively,
Because BD is an internal bisector
Therefore, ABF = NBF
Because AF ⊥ BD
Therefore, AFB = NBF = 90 degree
And because BF = BF
So △ Abf ≌ △ NBF
So AF = NF, ab = BN
Similarly, it can be proved that Ag = mg, AC = cm
So FG is the median of △ amn
So FG = Mn / 2
Because Mn = bc-bm-cn
That is Mn = BC - (bn-mn) - (cm-mn)
The results show that Mn = AB + ac-bc
So FG = (AB + ac-bc) / 2
2、
F and N are the same as above
Similarly, f is the midpoint of an, ab = BN
In addition, it can also be proved by congruent triangle
G is the midpoint of am, AC = cm
Because CE is the outer bisector of ABC
So the difference from the above question is that M is on the extension line of BC
Mn = BC + cm-bn = BC + ac-ab
So FG = (BC + ac-ab) / 2

It is known that, as shown in the figure, BD and CE are bisectors of the outer angle of triangle ABC respectively. AF is vertical BD through point a, and Ag is vertical CE
FG = (1 / 2) (AB + BC + AC),

Extend AF, extend CB to H. extend AG, extend BC to K. ∵ BD, divide ABC, ≌ △ Abf ≌ HBF.AF=FH . AB = Hg. ∵ CE bisection ∠ ACK, ≌ △ ACG ≌ △ KCG.AG=GK In addition, ab = Hb, AC = KC, BC = BC, HK = AB + BC + AC

BD and CE are bisectors of the exterior angle of triangle ABC respectively. AG is vertical to CE, AF is vertical to BD and FG. What is the relationship between BD and FG and the three sides of triangle ABC
BD and CE are bisectors of triangle ABC, Ag vertical CE, AF vertical BD,
Syndrome: FG = (BC + ac-ab)

The height is outside the triangle
I think FG = 0.5 (AB + ac-bc)
Extend BC, extend AF to h, extend CB, extend AG to I
△ABF≌△HBF,△ACG≌△ICG(SAS)
AF=FH,AG=GI
Ψ GF = 0.5ih (median line)
And IH = BH + ic-bc
∴GF=0.5(AB+AC-BC)
The height is in the triangle
Lengthen AG, cross BC to h, lengthen AF, cross BC to I
△ABF≌△IBF,△ACG≌△HCG(SAS)
AB=IB,CA=CH
Ψ GF = 0.5hi (median line)
∴GF=0.5(AB+AC-CB)

In RT △ ABC, angle c = 90 degrees, AC = 3cm, BC = 4cm, draw a circle with point C as the center. When the circle C has the following position relationship with line AB, try to find the value range of radius r of circle C
(1) Circle C is separated from ab;
(2) Circle C is tangent to ab;
(3) Circle C intersects line ab

When passing through the point C, make CD ⊥ AB in D ∵ - C = 90, AC = 3, BC = 4 ≁ AB = √ (AC + BC) = √ (9 + 16) = 5 ∵ CD ⊥ ab ≁ s △ ABC = CD × AB / 2 ∵ - C = 90 ∫ s △ ABC = AC × BC / 2 ∫ ad × AB = AC × BC ∫ ad × 5 = 3 × 4 ∫ ad = 2.4 ∫ when R < 2.4, circle C is separated from AB; when r = 2.4, circle C is separated from

It is known that the line y = K (x-m) and the parabola y2 = 2px (P > 0) intersect at two points a and B, and OA ⊥ ob, OD ⊥ AB are at point D. if the coordinates of the moving point d satisfy the equation x2 + y2-4x = 0, then M is equal to ()
A. 1B. 2C. 3D. 4

Let D coordinate be (x, K (x-m)), then the slope of OD is k ′ = K (x − m) x; let D coordinate be (x, K (x-m)), then the slope of OD is k ′ = K (x − m) x; let OD ⊥ ab, the slope of AB is k ⊥ K · K ′ = K2 (x − m) x = - 1, that is, K (x-m) = - XK; let D coordinate satisfy x2 + y2-4x = 0, that is, X2 + [K (x-m)] 2-4x = 0

It is known that the square of the line and the parabola y = 2px (P greater than 0) intersects two points a and B, and OA is perpendicular to ob, OD is perpendicular to AB, intersecting AB at D, D coordinates (3, √ 3)
Given that the square of the line and the parabola y = 2px (P greater than 0) intersect two points a and B, and OA is perpendicular to ob, OD is perpendicular to AB, intersecting AB at D, D coordinates (3, √ 3), calculate P

The equation of OD is y = √ 3 / 3x, so the slope of line AB is √ 3, so the equation is Y - √ 3 = √ 3 (x-3), so y = √ 3x-2 √ 3
Substituting Y & # 178; = 2px, we get 3x & # 178; - (12 + 2P) x + 12 = 0
So x1x2 = 4
So (y1y2) &# 178; = 4P & # 178; x1x2 = 16p & # 178;, so y1y2 = - 4P
Because OA ⊥ ob, so x1x2 + y1y2 = 0, that is 4-4p = 0, so p = 1

It is known that a straight line intersects a parabola y & sup2; = 2px (P > 0) at two points a and B, and OA ⊥ ob, OD ⊥ AB intersects AB with point D
(1) Finding the trajectory equation of point d
(2) If the coordinate of point D is (2,1), find the value of P

Let a (a ^ 2 / (2P), a), B (b ^ 2 / (2P), b), D (x, y) OA ⊥ oba / (a ^ 2 / (2P) * B / (b ^ 2 / (2P) = - 1ab = - 4P ^ 2 od ⊥ aby / X * (a-b) / [(a ^ 2-B ^ 2) / (2P)] = - 1y / x = - (a + b) / (2P) point D on the straight line AB, Y-A = (a-b) / [(a ^ 2-B ^ 2) / (2P)] [(x-a ^ 2 / (2P)] (a + b) (Y-A) = 2px-a ^ 2 (a +

It is known that a (x1, Y1), B (X2, Y2) are two points on the parabola y & # 178; = 2px (P > 0), satisfying OA ⊥ ob, O as the coordinate origin
Verification: the line where AB is passing the fixed point (2P, 0)

Let AB and X axis intersect at t (m, 0) (m ≠ 0)
Then the equation of line AB can be set as
x=ty+m
X = ty + m and Y & # 178; = 2px eliminate x simultaneously
y²=2pty+2pm
That is Y & # 178; - 2pty-2pm = 0
According to Weida's theorem
y1+y2=2pty,y1y2=-2pm
∴x1x2=(y²1/2p)*(y²2/2p)²
=(y1y2)²/(4p²)=m²
∵OA⊥OB
The vector OA ● ob = (x1, Y1) ● X2, y2 = 0
∴x1x2+y1y2=0
∴m²-2pm=0
∵m≠0,∴m=2p
That is, the line AB passes through the fixed point (2P, 0)

Let a (x1, Y1), B (X2, Y2) be two points on the parabola y2 = 2px (P > 0) and satisfy OA ⊥ ob, then y1y2 is equal to______ ．

∵ a (x1, Y1), B (X2, Y2) are two points on the parabola y2 = 2px (P > 0), and satisfy OA ⊥ ob. ∵ KOA · kob = - 1, ∵ x1x2 + y1y2 = 0, ∵ (y1y2) 24p2 + y1y2 = 0, then y1y2 = - 4p2, so the answer is: - 4p2

Given that the line passing through the point m (2P, 0) intersects the parabola y & # 178; = 2px (P > 0) and two points AB, we prove OA ⊥ ob
When the line AB is at what position, the area of △ AOB is the smallest