As shown in the picture, there is a section on both sides of a river that is parallel. There is a tree every 5 meters on the South Bank of the river, and a telegraph pole every 50 meters on the north bank. Xiaoli stands at point P 15 meters away from the south bank and looks at the North bank. She finds that two adjacent telegraph poles A and B on the north bank are just covered by two trees C and D on the south bank, and there are three trees between the two trees

As shown in the picture, there is a section on both sides of a river that is parallel. There is a tree every 5 meters on the South Bank of the river, and a telegraph pole every 50 meters on the north bank. Xiaoli stands at point P 15 meters away from the south bank and looks at the North bank. She finds that two adjacent telegraph poles A and B on the north bank are just covered by two trees C and D on the south bank, and there are three trees between the two trees

Let p be PF ⊥ AB, intersect CD at e, intersect AB at F, as shown in the figure: let the river width be x M. ∵ ab ∥ CD, ∵ PDC = ∵ PBF, ∵ PCD = ∵ PAB, ∵ PDC ∵ PBA, ∵ ABCD = PFPE, ∵ ABCD = 15 + x15, CD = 20 m, ab = 50 m, ∵ 2050 = 1515 + X according to the theme, the solution is: x = 22.5 (m). Answer: the river width is 22.5 M

As shown in the picture, there is a section on both sides of a river that is parallel. There is a tree every 5 meters on the South Bank of the river, and a telegraph pole every 50 meters on the north bank. Xiaoli stands at point P 15 meters away from the south bank and looks at the North bank. She finds that two adjacent telegraph poles A and B on the north bank are just covered by two trees C and D on the south bank, and there are three trees between the two trees

Let p be PF ⊥ AB, intersect CD at e, intersect AB at F, as shown in the figure: let the river width be x M. ∵ ab ∥ CD, ∵ PDC = ∥ PBF, ∥ PCD = ∥ PAB, ∥ PDC ∥ PBA, ∥ ABCD = PFPE, ∥ ABCD = 15 + x15, CD = 20 m, ab = 50 m, ∥ 2050 = 1515 + X according to the theme, the solution is: x = 22.5 (m)

As shown in the picture, a and B are on both sides of the same river. Now we need to build a bridge Mn on the river. Where can we make the path from a to B the shortest? In the figure below, draw the path, not the drawing method, but explain the reason

As shown in the figure, make BB 'perpendicular to the river bank GH, make BB' equal to the river width, connect ab ', intersect with the river bank EF at m, make Mn ⊥ GH, then Mn ∥ BB' and Mn = BB ', so mnbb' is a parallelogram, so NB = MB '. According to "the shortest line between two points", ab' is the shortest, that is, am + BN is the shortest. So the bridge is built at Mn

The two sides of the river are parallel. A and B are the two workshops on both sides of the river, as shown in the picture The question is too long. The rest is in the supplement
The two banks of the river form parallel lines. A and B are the two workshops located on both banks of the river (as shown in the figure). To build a bridge on the river, make the bridge perpendicular to the river bank, and make the distance between a and B the shortest. The method to determine the position of the bridge is as follows: make a vertical line from a to the river bank, and cross PQ and Mn to f and g respectively. Take AE = FG on Ag, and connect them EB.EB Cross Mn to D. make a vertical line DC at d to the opposite bank, Then DC is the location of the bridge. Try to say the shortest route when the bridge is built at CD, that is, the reason why (AC + CD + dB) is the shortest

Known, AE ⊥ PQ, CD ⊥ PQ, can be: AE ⊥ CD
Because, AE ‖ CD, AE = FG = CD,
Therefore, AEDC is a parallelogram and AC = ed
CD is a fixed value. To make AC + CD + DB shortest, AC + DB shortest
Because AC + DB = ed + DB ≥ EB,
Therefore, AC + CD + DB = CD + (AC + dB) ≥ CD + EB;
Among them, the equal sign holds when e, D and B are collinear (between two points, the line segment is the shortest)
When point D is the intersection of EB and Mn, AC + CD + DB is the shortest

There are two parallel lines on both sides of the river. There are two villages a and B on one side of the river. A bridge should be built on both sides of the river
There are two parallel lines on both sides of the river. There are two villages a and B on one side of the river. It is necessary to build a bridge on both sides of the river (the bridge deck is perpendicular to the river bank). Where is the bridge built to make the sum of the distance between the two villages and the bridgehead shortest? Why
It's better to attach a picture

Is the village on the same side

As shown in the figure, parallel lines AB and CD are cut by straight line AE
(1) How many degrees is ∠ 2? Why?
(2) How many degrees is ∠ 3? Why?
(3) How many degrees is ∠ 4? Why?

1 and 2 are stagger angles
If parallel, the internal stagger angles are equal
So ∠ 2 = 110 degrees
1 and 3 are appositive angles
Parallel is equal to the same angle
So ∠ 3 = 110 degrees
1 and 4 are ipsilateral angles
Parallel is complementary to the internal angle of the same side
Therefore, 4 = 180-110 = 70 degrees

It is known that there are two villages a and B on both sides of the river. Now we need to build a bridge to make the bridge perpendicular to the river bank and the shortest distance. Where should the bridge be built?

A bridge perpendicular to the river will be built outside the junction of AB village and the river

On a parallel river bank, build a bridge perpendicular to the river bank. Where can the bridge be built to minimize the distance between two fixed points on both sides of the river bank?
Maybe you don't understand the meaning of my question. That is to say, the river banks are parallel, and the bridge must be vertical to the two banks, forming an I-shape. There are two fixed points (A and b) on both sides of the river. Ask where the bridge is built to make the shortest distance from point a to point B

The bridge is in a straight line A
Two points on both sides of the river bank form a straight line B
Point C is the point where the midpoint of line a and line B coincide at the center of the river
Build a bridge perpendicular to the river bank at point C
The shortest distance between the bright spots on the river bank

As shown in the figure, it is known that Mn is the vertical bisector of BC side of △ ABC and intersects AB at P, ab = 9, AC = 8. Try to find the perimeter of △ PAC

⊿ perimeter of PAC = PA + PC + AC
Mn is the vertical bisector of BC
∴PB=PC
∴PA+PC=PA+PB=AB
The perimeter of PAC = AB + AC = 9 + 8 = 17

As shown in the figure, in △ ABC, the bisector of ∠ ABC and ∠ ACB intersects at point O, passing through point o as Mn ‖ BC, respectively intersecting AB and AC at points m and N. if AB = 12, AC = 18 and BC = 24, then the perimeter of △ amn is ()
A. 30B. 36C. 39D. 42

As shown in the figure, ∵ OB and OC are bisectors of ∵ ABC and ∵ ACB, ∵ Mn ∥ BC, ∵ Mn ∥ BC, ∵ Mn ∥ BC, ∵ Mn = ma + an + Mn = ma + an + Mo + on = AB + AC, and