The correctness of convex lens imaging formula is proved by junior high school geometry knowledge,

The correctness of convex lens imaging formula is proved by junior high school geometry knowledge,

It can be proved by the similarity of two groups of right triangles that the heights of the images are equal
Verification: the proof of plane geometry in junior high school
Shame, shame. I was choked by my nephew in junior high school. Everyone, who still remember the plane geometry theorem in junior high school? Please help me to give the following proof process. Thank you
m(_ _ M

BCE




AE and BF intersect at d 
Known:  ab ⊥ BC            
FE⊥BC
∠ACB=∠ECFD

Verification:  DC ⊥ BC

A






F
Proof: ∵ ab ⊥ be EF ⊥ be
The AB is parallel to ef
The Δ abd is similar to the Δ EFD
∴BD/FD=AB/EF
∵∠ACB=∠FCE ∠ABC=∠FEC
The ABC is similar to the FEC
∴BC/EC=BA/EF=BD/FD
The Δ BCD is similar to the Δ bef
∴∠BCD=∠BEF=90°
∴DC⊥BC
Junior high school geometry proof questions,
The two heights of triangle ABC are be and cf. point m is the midpoint of BC and point n is the midpoint of EF,
Verification of Mn ⊥ ef
If the center line of the hypotenuse of the right triangle BFC is equal to half of the hypotenuse, then MF = 1 / 2BC
So me = MF
Do you understand? I'll teach you if you don't understand
On geometry proof of junior high school
Isosceles right triangle ABC, a is a right angle, M is the midpoint of AC side, connecting BM, ad is perpendicular to BM at e point
Verification ∠ AMB = ∠ DMC
The answer is fast and accurate, give 50 points more!
yes,
It is proved that if an ⊥ BC intersects at point n and BM intersects at point P, ⊥ ban = ⊥ can = ⊥ ABC = ⊥ C = 45 °
AN=BN=CN AB=AC ∠DAC=∠ABM=90°-∠BAE
The value of ∧ ACD is equal to ∧ ABP (SAS) ∧ AP = CD and am = MC ∧ amp ≡ CMD (SAS)
∴∠AMB=∠DMC
Is d on BC
As shown in the figure, in the trapezoidal ABCD, ab ‖ CD, the median line EF intersects the diagonal lines AC and BD at two points m and N. if EF = 18cm, Mn = 8cm, then the length of AB is equal to______ cm．
∵ EF is the median line of trapezoid, and EF = 18cm, ∥ AB + CD = 2 × 18 = 36cm, EF ∥ ab ∥ CD. ∥ am = cm, BN = DN. ∥ EM = NF = 12CD = 18 − 82 = 5. ∥ CD = 10 ∥ AB = 2ef-cd = 36-10 = 26 (CM)
First question: known circle C: x2 + (Y-1) 2 = 5, straight line L: mx-y + 1-m = 0
If the fixed point P (1,1) chord AB is AP / Pb = 1 / 2, the equation of direct L is obtained
Question 2: angle a is 60 degrees, if B + C = root 3 * a, try to judge the shape of the triangle
Question 3: given that two of the equations x2 + (2 + a) x + 1 + A + B = 0 are x1, X2, and 0 is less than x1, X1 is less than 1, X2 is greater than 1, then the value range of B / A
The fourth question: when the electric fan hanging on the ceiling turns on, the pulling force on the ceiling is larger than that when the electric fan turns on. A: when it turns on, B: when it turns on, B: when it turns on, C: when it turns on, D: it's not sure
Question 5: the plane carrying out the disaster relief mission flies in a straight line at a horizontal and uniform speed against the wind, and releases two boxes of disaster relief materials 1 and 2 with exactly the same shape and quality at an interval of 0.5 seconds, People on the ground look along the direction of the plane. A: box 1 is directly below box 2. B: the horizontal distance between the two boxes remains unchanged. C: the horizontal distance between the two boxes is getting larger and larger. D: the horizontal distance between the two boxes is getting smaller and smaller
Question 6: an adult and a child stand on the horizontal ground, hand in hand to compare their strength. As a result, the adult pulls the child over
A: Adults pull children more than children pull adults
B: The tension between adults and children is a pair of balancing forces
C: The force that an adult pulls a child must be equal to the force that a child pulls an adult
D: Only in the process of pulling the child by adults can the force of adults be greater than that of children. In the possible short-term stalemate, the pulling force of two people is the same
Note: I know all the answers to these questions, so you know I need to solve them in great detail. I'm sorry for wasting your precious time. But if you want to complete the answers, and don't read the wrong questions, then 30 points will be yours. Ha ha, this is all my property. I've been playing for half an hour. I hope you can give me a good answer. Thank you
The score is high. Please don't answer like the one below. I'll give you all the marks
More questions, less marks
In the first question, AB should be the intersection of L and circle, but if not, it can't be solved
First find the coordinates of the intersection point, and then calculate them with the distance formula, and then calculate them with AP / BP = 1 / 2
The fourth question should be that there is no comparison between rotation and rotation. When rotating, there is wind support, so the tension on the ceiling becomes smaller
There is motion, but the pulling force is greater than the ground friction
Write directly on the answer sheet: Baidu, you will know. And draw bear's paw sign, I believe you can get full marks!
The answer to question 6 should be B. I think the role of force is mutual, so it should be
Question 4: when the fan rotates, there is air flow downward, so choose B
Question 5: D
Question 6: C force and reaction
A few high school mathematics + physics problems!
1. Set M = {m > = - 1 / 4 or M = 0}, set n = {n}
1. It's equal to M
2. Meter ruler
How long does it take for each drop
And then walk a certain distance, a total of a few drops of liquid
1`CuM={X
It is known that f (x) is a function over R, and f (x + 2) = (1 + F (x)) / (1-f (x)). It is proved that f (x) is a periodic function
A plane moves upward with uniform acceleration along the 30 ° elevation direction, and the acceleration is 10m / s and 178; after 4 S, what is the displacement of the plane? How much does the plane rise in the vertical direction?
f(x+2)=(1+f(x))/(1-f(x))
f(x+4)=(1+f(x+2))/(1-f(x+2))=-1/f(x)
f(x+6)=(1+f(x+4))/(1-f(x+4))=(1+f(x))/(1-f(x))=f(x+2)
Because f (x + 2) = f (x + 6)
So f (x) is a function of period = 4
When a car moves in a straight line on a horizontal road, its power remains unchanged. When the speed of the car is 4m / s and the acceleration is the square of 0.4m/s, the resistance of the car is always 0.01 times of the weight of the car. If G is taken as 10, the maximum speed of the car can be calculated
The mass m = 1.5kg can be regarded as a mass. Under the action of horizontal constant force F, the mass starts to move from static point a on the horizontal plane, and the force is removed for a certain distance. The mass continues to slide for 2S and stops at point B. It is known that the distance between two points a and B is 5m, and the dynamic friction coefficient between the mass and the horizontal plane is 0.2
A shot with a mass of 2kg falls freely from 2m above the ground and sinks into a bunker 2cm deep. The average resistance of the sand to the shot is calculated
In triangle ABC, Tana: tanb = (root 2 times C) minus B: B, then angle a equals ()
A. 30 degrees B 45 degrees C 60 degrees
In the triangle ABC, it is known that ab = 2, ∠ C = 50 ° and when ∠ B is equal to what, BC gets the maximum value
In the acute triangle ABC, if ∠ C = 2 ∠ B, then the value range of AB: AC is--------
In triangle ABC, ABC is the opposite side of angle ABC, 4 times sin ^ 2 [(B + C) / 2] - con ^ 2A = 3.5
1. Find the angle of angle A
2. If a = radical three, B + C = 3, find the value of B and C
In the triangle ABC, if (a + B + C) (a-b + C) = 2Ac, and Tana + Tanc = 3 + radical 3, the height of side AB is 4 times the radical 3, the size of angle a, B, C and the length of side a, B, C are calculated
At the coast a, a smuggling ship was found at 45 ° n by E, at B [(root 3) - 1] km away from A. at 75 ° n by W, at C, 2km away from a, the anti smuggling ship was ordered to chase the smuggling ship at a speed of 3km / h at 10 root. At this time, the smuggling ship was fleeing from B to 30 ° n by e at a speed of 10km / h
Good answers can contribute to my existing scores
1. When the vehicle reaches Vmax, f = f, P = vmaxf
The results show that Vmax = P / F = 2m / 0.1M = 20m / s
2. The second process: a = f / M = (FN * U) / M = (15 * 0.2) / 1.5 = 2m / S ^ 2
v0=v-at=0m/s-(-2m/s^2)*2s=4m/s
x=v0t+1/2at^2=4m/s*2+1/2*(-2)m/s^2*(2s)^2=4m
The first process: F = 3N, a = (f-3n) / 1.5kg
v^2-（v0)^2=2ax
4^2-0^2=2*(F-3N)/1.5kg*1m
F=15N
3.1 / 2GT ^ 2 = 2m t = root 10 / 5
V = GT = 2 radical 10
v^2-（v0)^2=2ax
(2 radical 10) ^ 2-0 ^ 2 = 2 * a = 0.02M
a=1000m/s^2
F=ma=2kg*1000m/s^2=2000N
4.B
5. When B = 40 degree
How to memorize the physical formula quickly?
In the process of learning, we must: listen more, remember more, read more, do more homework, ask more, and do more experiments
If you understand him, you will remember him