Ask for the second year of junior high school math there is a bounty The second year of junior hig

p is the inner part of the triangle, which is the intersection of the bisector of the three angles.

Let the distance be x

s△abc = 1/2×ab×x + 1/2×bc×x + 1/2 ×ac×x

These are three small triangles, each distance is perpendicular)

25 24 7 is the Pythagorean number, which is a right triangle.

s△abc = 7×24×1/2=84

x=3 I've just done this question, and I'm just reviewing it, hehe.

p is equal to the distance to each side.

Explain that p is the inner part of the triangle and is set to r

1/2ab*r+1/2bc*r+1/2ac*r=s△abc1/2(25+24+7)r=s△abc

28r=s△abc

Let p=(ab+bc+ac) 2=28

s△abc=√[p(p-ab)(p-ac)(p-bc)]=√(28*3*4*21)=84

28r=84

r=3cm

It is known that P is the triangle of the heart.

Let the distance be r

s△abc = 1/2 ×ab×r + 1/2 ×bc×r + 1/2 ×ac×r

25 24 7 is a right triangle.

s△abc = 84

r = 3

p is the intersection of the perpendicular line.

Because the title says that the point p is a point in the triangle abc, the point p is the intersection of the three angle bisectors of the triangle abc (according to the distance between the two sides of the angle on the angle bisector is equal to the point on the angle bisector line), so the distance of the point ab can be completed

The triangle ABC is a right triangle.

Figure 2 should be the outer corner, right?

The answer is. In Figure 1, the intersection is the heart of the triangle. In Figure 2, the intersection point is the side.

The nature of the heart, the side mind, is that the distance to the three sides is equal.

So there are 4 points in question three. One heart, three side hearts.

It should not be required to master the specific proof of intersecting junior high school, but it can be proved in a variety of ways.

The proof upstairs is too sketchy.

Give a more formal and common proof method:

It is known that in the triangle ABC, the two angular bisectors BM and CN intersect at P

Verification: The point p is on the bisector of the angle bac, and the distance from the point p to the three sides is equal.

Proof: PE is perpendicular to E, PD is perpendicular to Ab to D, PF is perpendicular to Ac to F

BM bisects the angle ABC, then PE=PD; CN bisects the angle ACB, then PE=PF(properties of angular bisector).

Therefore PE=PD=PF.The point p is on the bisector of the corner bac.

1. Yes, because the three angular bisectors intersect at one point, the intersection of the two angular bisectors must be on the angular bisector of c, and the proof method is: pass the point o as a perpendicular line to cross bc ac at m, n. Because om=on, the point o is on the angular bisector of c.

2. Yes, ditto.

3. There is only one, and the distance from the point on the bisector of the angle to both sides of the angle is equal.

1) Solution: o is on the perpendicular bisector of c.

Because: o is on the angular bisector of a and b.

So: o on the angular bisector of c (the angular bisector of the triangle intersects at one point).

(1) From the question, we know that 2x +7x-1=4x+1

2x²+3x-2=0

2x-1)(x+2)=0

2x-1=0 or x+2=0

x = 1 2 or -2

2) From the question, we know that 2x +7x-1+(-17-x)=0x +7x-18=0

x+9)(x-2)=0

x+9=0 or x-2=0

x=-9 or x=2

The line segment mn=on, mn x-axis, δomn is an isosceles right triangle, mon=45°

If F sits on Fg parallel BC, then the triangle BED is all equal to the triangle FGD, BE=FG, and the triangle AFG is equal to the triangle AEC, then BE:EC and other FG:EC=AF:AC=1:3

Do AE parallel lines Fm cross EC for M points.

It is possible to obtain em:mc=1:2 and be:em=1:1

So be:ec=1:3

Solution: Parallel lines of bf are crossed by d, and the AC edge is at g, as shown in the figure below

D is the midpoint of BC, and DG BF

cgd=∠cfb

again c= c

cdg∽△cbf

CG CF= cd CB= 1 2, that is: CG= 1 2CF=FG and E is the midpoint of AD, and the extension line of BE intersects AC at F, and DG BF can be obtained in the same way: AEF ADG

ae ad= af ag= 1 2, i.e.: af = 1 2ag = fg af = fg = gc

af fc= af 2af= 1 2=1:2, so d

(1) Majority, the most people, 240

Median, highest and lowest mean = (540 + 120) 2 = 330 average = (540x1 + 450x1 + 300x2 + 240x6 + 210x3 + 120x2) (1 + 1 + 2 + 6 + 3 + 2) = 3900 15 = 260

2) 260 is the average, but most people can't do it, it's unreasonable.