Junior 3 Mathematics A difficult geometry problem 20

This problem is not easy for me to do, you can think about it again, I do it in two steps, even pb to make a small circle to m, even ap, qb

The first step is to prove that RM is parallel to QB, and I didn't come up with a good way to do this, which is to connect PO!o Use similar proportions to prove, this step is no problem, but it depends on what method you can use to prove it! I don't know what you've learned!

The second step, the previous step is to get the angle PMR angle pbq angle pra = angle PMR angle brq angle pbq (chord cut angle, to the apex angle.

So, the angle brq angle pbq

So, the angle abq angle bpq (similar triangle, or, two triangles have two corresponding angles that are equal and the last angle is also equal).

So q is the midpoint of the arc ab.

I do the perpendicular line of the bottom. Construct two right-angled triangles, because one angle is 30 degrees, so the side corresponding to the 30-degree angle is half of the hypotenuse, that is, 1. According to the Pythagorean theorem, the length of the base is the root number (2 squared minus 1 squared) is the root number 3.

Therefore, the length of the bottom edge is 2 and 3. The area of the triangle is the base multiplied by the height = 2 root number 3 * 1 2 = root number 3. Landlord, I wrote it right.

The adjacent edge of the cos cos angle hypotenuse (note: adjacent edge - right-angled edge).

The opposite edge of the positive angle of the sin is the hypotenuse (note: the opposite side - the right angle).

tan tangent of the opposite side of the adjacent edge (note: ratio of two right-angled edges).

The adjacent edge of the cot tangent angle is opposite the edge (note: the ratio of two right-angled edges).

Special value: cos0°=1;cos30°=(root number 3) 2;cos45°=(root number2) 2;cos60°=1/2

cos90°=0

sin0°=0；sin30°=1/2；sin45°=(root number2) 2;sin60°=(root number 3) 2;sin90°=1

tan0°=0；tan30°=(root number 3) 3;tan45°=1；tan60° = root number 3;

tan90°=infinity;

cot0°=infinity; cot30° = root number 3; cot45°=1；cot60°=(root number 3) 3;

cot90°=0；

Make a perpendicular line (bottom) and then do a good pull

Isn't it cos, sin, and tan? , look up a table, or get a calculator!

From the top angle to the enemy, it becomes a perpendicular line, isn't it a right triangle of two townships? ，

ce=df+cg

After D is the perpendicular line of DE, the extension line of EF is crossed with H, which proves that EDH is equal to ABG, and then proves that DF=DH, so that DF=BG=BG=CE

In the proof upstairs: de=ad=ec, this is problematic, how to prove ad=de?

Over e for EM vertical AD, m for the vertical foot.

de=2em

em＝x2x）²－x²＝（a／2）²

3x²＝a²／4

Find: It is easy to get ae=de=ef

So what is needed is:

Extend DE to AB and G, Extend EF to AD to H, and BC to I.

Angle AEG = EAD + EDA = 30 + 30 = 60 [Outer Angle Theorem], Angle EAG=90-EAD=60

Angle ega = 180-aeg-eag = 60 [eag inner angle and 180].

EAG is an equilateral triangle [60 at all three corners].

ag=ae=eg=ed [The triangle EAD is an isosceles triangle, AE=ED].

In the right-angled triangle GAD, let ag=b, dg=2b, and the Pythagorean theorem obtains, AD=root number3B=A, B=A, root number3

ae+ed=2a root number 3, the same way bf+cf = 2a root number 3

eh=fi [easy to prove congruence, omitted] e is the midpoint of eg, eh is the median of the triangle dag, eh=1 2ag [median line theorem].

So eh+fi=2eh=ag

ef=ab-fi-eh=ab-ag=a-b

Total wire length = a-b + 2a root number 3 + 2a root number 3 = a + 3a root number 3 = a + (root number 3) a

Note: Add and supplement by yourself].

Let de=x, ad=root number 3x (pass e to make a perpendicular line, 30 angles rt triangle, this you must have solved in junior high school).

Therefore x = root number 3a 3

Next, all kinds of perpendicular lines can be added to get ef=a(1-root number 3 3).

Feel free to ask.

E is EM vertical AD, m is perpendicular foot, F is fn vertical BC, n is perpendicular foot and EM x ade= aed=30

de=2em

It is obtained according to the Pythagorean theorem.

2x)²-x²=(a/2)²

3x²=a²/4

x=√3a/6

ef=(1-√3/3)a

The required wire length is (5-3 3)a

Over e for EM vertical AD, m for the vertical foot.

de=2em

em＝x2x）²－x²＝（a／2）²

3x a 4 x a root number 3 68x a 2 4a root number 3 3 a 2

The triangle ADE is an isosceles triangle, the AD length is A, the midline length is 3A 6, and the AE is 3A 3, and the same way for the triangle BCF, the EF length is (1- 3 3)A, so EF+4*AE is required

ae=ed=bf=cf=a/2

EP is the perpendicular line of AD, and FO is the perpendicular line of BC.

ep=fo=√(ae)^2-(ap)^2

ef=a-ep-fo=2√3