Take three points ABC on the line, such that ab = 8 cm, BC = 6 cm, if M is the midpoint of line AB, if n is the midpoint of line BC, Then the length of segment Mn is several centimeters

Take three points ABC on the line, such that ab = 8 cm, BC = 6 cm, if M is the midpoint of line AB, if n is the midpoint of line BC, Then the length of segment Mn is several centimeters

1 if C is between ab
M is ab, the midpoint AB = 12, so MB = 6
N is the midpoint of BC, BC = 8, so NB = 4, so Mn = mb-nb = 2
2 if C is outside ab
M is the midpoint of AB, ab = 12, MB = 6
N is the midpoint of BC, BC = 8, Nb = 4
MN=MB+NB=6+4=10
So Mn = 2 or Mn = 10
I hope I can help you!

As shown in the figure, M is the midpoint of AB, n is the midpoint of BC, and D is the midpoint of AC. if AC = 10cm, what is the length of Mn? Is Mn equal to DC?

No picture,
(1) It could be
A|--------M-----D----B---N----|C
AC=10
AD=CD=1/2AC=5
AB+BC=AC=10
1/2AB+1/2BC=1/2(AB+BC)=MN
∴MN=1/2AC=5
∴MN=CD
(2) Maybe
A------------D------M------C----N----B
It won't work out

Point C is on line ab. m and N are the midpoint of line AC and BC respectively. If Mn = 10cm, how many centimeters is ab equal to

AB = 2Mn = 20cm, draw the figure below, you can see,

Given that a, B and C are three sides of △ ABC, and A2 + B2 + C2 = AB + AC + BC, then △ ABC is ()
A. Isosceles triangle B. right triangle C. equilateral triangle D. isosceles right triangle

The original formula can be reduced to 2A2 + 2B2 + 2c2 = 2Ab + 2Ac + 2BC, that is, A2 + B2 + C2 + A2 + B2 + c2-2ab-2ac-2bc = 0; according to the complete square formula, we can get: (a-b) 2 + (C-A) 2 + (B-C) 2 = 0; from the properties of non negative numbers, we can know that A-B = 0, C-A = 0, B-C = 0; that is, a = b = C. so △ ABC is an equilateral triangle, so we choose C

Given that the three vertices of the triangle ABC are a (- 2,2), B (1,4), C (5,2), the equation for finding the circumscribed circle of the triangle is

The distance from the center of a circle to (- 2,2), (5,2) is equal, so it is necessary to set the center of a circle as (3 / 2, b) radius r on the straight line x = (- 2 + 5) / 2 = 3 / 2, then the circular equation (x-3 / 2) &# 178; + (y-b) &# 178; = R & # 178; (5,2), (1,4) is substituted by (5-3 / 2) &# 178; + (2-B) &# 178; = R & # 178; (1-3 / 2) &# 178; + (4-b) &# 178; = R & # 178

Known: ABC vertex coordinates a (5, - 3), B (6,2), C (- 4,4), let D, e, f be the midpoint of AB, BC, CA respectively, try to find D

D (11 / 2, - 1 / 2), because a (5, - 3), B (6,2), D is the midpoint of a and B, so the coordinates of point D are (5 + 6) / 2, (- 3 + 2) / 2, so the coordinates of point D are (11 / 2, - 1 / 2)

It is known that in △ ABC and △ def, ab = De, am and DN are the heights of △ ABC and △ def respectively, am = DN. Try to explore the relationship between ∠ ABC and ∠ def, and explain the reason

① ∵ am and DN are the heights on the sides of △ ABC and △ def, respectively, ∵ AMB = ∠ dne = 90 ° and ∵ AB = De, am = DN, ≌ ABM ≌ den (HL), ≌ can get ≌ ABC = ∠ def. ② ∵ AB = De, am = DN, ≌ ABM ≌ den (HL), ≌ can get ≌ ABC and ≌ def complementary

It is known that: in △ ABC and △ def, ab = De, BC = EF, am is the middle line of △ ABC, DN is the middle line of △ def, am = DN, the proof is △ ABC ≌ Δ def

It is proved that: ∵ BC = EF, am is the middle line of △ ABC, DN is the middle line of △ def, ∵ BM = en, in △ ABM and △ den, ab = Adam = dnbm = en ≌ ABM ≌ Δ den, ≌ B = ≌ e, in △ ABC and △ def, ab = de ≌ B = ≌ EBC = EF ≌ ABC ≌ def

It is known that am is the midline of △ ABC, and DN ‖ am intersects AC with E. to verify AD / AE = AB / AC

N is on BC, D is on BA
∵AM∥DN,
The results show that AB / ad = BM / Mn, AC / AE = cm / Mn (parallel line segments are proportional),
∵ m is the midpoint of BC, ∵ AB / ad = cm / Mn,
∴AB/AD=AC/AE,
{AD / AE = AB / AC (internal exchange)

Triangle ABC is equal to triangle def, am is perpendicular to BC, DN is perpendicular to EF, am and DN are equal? Please explain the reason
It's two separate acute triangles,

Congruent triangles have the same area and the same side length
And 1 / 2am * BC = 1 / 2DN * EF
Am = dn