As shown in the figure, the turning point of a highway is an arc. The point O is the center of the arc, AB is equal to 120m, C is the point on AB, O is the point on AB, the perpendicular foot is D, and CD is equal to 20m. Calculate the radius of the curve

As shown in the figure, the turning point of a highway is an arc. The point O is the center of the arc, AB is equal to 120m, C is the point on AB, O is the point on AB, the perpendicular foot is D, and CD is equal to 20m. Calculate the radius of the curve

Let the radius be X
Find the center of the circle and draw a circle. Then we get the basic figure of the vertical longitude theorem
According to the relevant series of equations on the line. Hey, I just did this problem
Come on

As shown in the figure, after two turns of a highway, it is the same as the original direction. If the first turning is 36 ° and the second turning is how many degrees? Why?

According to the complementary inner angle of the same side, the angle of the second turning is 180 ° - 36 ° = 144 °

As shown in the figure, in the triangle ABC, D is the midpoint of BC, De is vertical AB, DF is vertical AC, points E and F are perpendicular feet, De is equal to DF It is proved that triangle bed is equal to triangle CFD

prove:
∵DE⊥AB,DF⊥AC
∴∠DEB＝∠DFC＝90
∵ D is the midpoint of BC
∴BD＝CD
∵DE＝DF
∴△BDE≌△CDF （HL）
∴∠B＝∠C
∴AB＝AC
Hope to solve your problem

It is known that in the triangle ABC, the angle BAC = 90, ad vertical BC, e is the midpoint of AC, and the extension line of ED intersecting AB is at F. it is proved that ab: AC = DF: AF

△ABD∽△BCA
There are AB / AC = BD / AD, ∠ bad = ∠ ACB
Because ad ⊥ BC, e is the midpoint of AC
So: De is the center line, ∠ EDC = ∠ BDF = ∠ ACB
So: ∠ bad = ∠ BDF, ∠ f is the common angle
So: △ FBD ∽ FDA
There are: DF / AF = BD / ad
So: AB / AC = DF / AF

As shown in the figure, it is known that in △ ABC, ∠ BAC = 90 ° ad ⊥ BC, e is the midpoint of AC, ed intersects AB extension line at F, and verification: ab AC=DF AF．

It is proved that: ? BAC = 90 °, ad ⊥ BC,  CBA ∵ abd,  abd = Acad,  AB: AC = BD: ad ①, ∵ C = ∵ fad, and ? e is the midpoint of AC, ad ⊥ BC, ≁ ed = 12ac = EC, ? C = ∧ EDC, and ∵ ∵ EDC = ∵ FDB, ? f is common angle,  DBF ? ADF

The triangle ABC, angle BAC = 90 degrees, ad perpendicular to D, e is the midpoint of AC, the extension of ED intersects the extension of AB at point F, and ab * AF = AC * DF Explain why

It is proved that ad is perpendicular to D and E is the midpoint of AC,
Therefore, de = EC = 1 / 2 * AC
Angle c = angle EDC
The angle BAC = 90 degrees, ad is perpendicular to BC and D,
So, angle c = angle bad
So, angle EDC = angle bad
Angle EDC = angle FDB
So, angle FDB = angle bad
Angle f = angle f
Therefore, triangle AFD is similar to triangle DBF
Therefore, AF / DF = ad / BD
Abd = angle abd
Angle bad = angle ACD
Therefore, triangle abd is similar to triangle CAD
Therefore, AC / AB = ad / BD
Therefore, AC / AB = AF / DF
Therefore, AB * AF = AC * DF

As shown in the figure, it is known that in △ ABC, ∠ BAC = 90 ° ad ⊥ BC, e is the midpoint of AC, ed intersects AB extension line at F, and verification: ab AC=DF AF．

It is proved that: ∵ BAC = 90 °, ad ⊥ BC,
∴△CBA∽△ABD，
∴AB
BD=AC
AD，
∴AB：AC=BD：AD①，
∴∠C=∠FAD，
And ∵ e is the midpoint of AC, ad ⊥ BC,
∴ED=1
2AC=EC，
∴∠C=∠EDC，
And ∵ EDC = ∵ FDB,
Ψ fad = ∠ FDB, ∠ f is the common angle,
∴△DBF∽△ADF，
∴BD：AD=DF：AF②，
AB is obtained from ① and ②
AC＝DF
AF．

In △ ABC, ∠ BAC = 90 °, ad ⊥ BC at D, e as the midpoint of AC, and de crossing Ba extension line at F. verification: ab: AC = BF: DF

Two knowledge points are used in this paper. One is that the center line on the hypotenuse of a right triangle is equal to half of the hypotenuse; the other is that the two right triangles divided by the vertical line on the hypotenuse of the right triangle are similar to the original right angle;
As shown in the figure:
If e is the midpoint of AC, then AE = de; angle ead = EDA; and because ad is a vertical line, so the triangle ADB, ACD, ABC are similar; then the angle ead = abd = ade;
Then the angle DBF = ADF, and the angle f is a common angle, then the triangle FBD and fad are similar;
Then BF: DF = BD: AB;
Because the triangle abd is similar to ABC, then BD: ab = AB: AC;
AB: AC = BF: DF
For the sake of drawing, typing and explaining clearly,

As shown in the figure, in the triangle ABC, the known angle a = 90 ° AB = AC, BD is the median line, AE is perpendicular to BD at e, and extended AE intersects BC at F. it is proved that angle ADB = angle CDF

Make the vertical lines of BC through a and D respectively, and the vertical feet are g and H
Let Ag = 1, then CG = 1, DH = 1 / 2, BH = 3 / 2
tang∠DBH=1/3
∠ GAF = ∠ DBH, so GF = Ag / 3 = 1 / 3
FH=GH-GF=1/2 -1/3 =1/6
tang∠FDH=FH/DH=1/3
So ∠ DBH = ∠ FDH
∠ADB=∠DBH+∠C
∠CDF==∠FDH+∠CDH
So: ∠ ADB = ∠ CDF

It is known that in △ ABC, ∠ a = 90 °, ab = AC, D is the midpoint of AC, AE ⊥ BD is at e, and extended AE is crossed with BC at F. it is proved that ∠ ADB = ∠ CDF

It is proved that if Ag = 1, then CG = 1, DH = 12, BH = 32, Tan ∠ DBH = 13, GAF = ∠ DBH, GF = 13ag = 13, FH = gh-gf = 12-13 = 16, Tan ∠ FDH = fhdh = 13 ﹤ DBH = ∠ FDH ? ADB = ∠ DBH + ∠ C, ∠ CDF = ∠ FDH + ∠ C