Junior high school geometry problem circle math master help to do it. Thank you! 150

Solution: bac=180°- abc- acb=50°Connect AI, BI, CI, AE

Point i is the heart of ABC, then: cai=(1 2) cab=25°, ACI=(1 2) ACB=30°

aic=180°-(cai+∠aci)=180°-(25°+30°)=125°;

ade=∠abc=70°;de=ad.

dea=∠dae=(180°-∠ade)/2=(180°-70°)/2=55°.

Then dea+ aic=180°Therefore, the four points a, i, c, and e are on the same circle.

dei=∠cai=25°.

It is known that E is a point on the extension line of the side CD of the circle bordered by the quadrilateral ABCD, and I is the heart of ABC, if ABC 70°, ACB 60°, de Da, then the degree of DEI is (25°).

Connect to AI CI

ABC. bac=180°-∠abc-∠acb=180°-70°-60°=50°

1=∠acb/2=30°

2=∠bac/2=25°

DAE. ed=ad

dea=∠dae

dea = dae = (180°-70°) 2 = 55 ° dae. AIC=180°- 1- 2=125°AIC+ DEA=125°+55°=180°AECI concomitation.

dei=∠2=25°

Point i is the heart of ABC, then: cai=(1 2) cab=25°, ACI=(1 2) ACB=30°

aic=180°-(cai+∠aci)=180°-(25°+30°)=125°;

ade=∠abc=70°;de=ad.

dea=∠dae=(180°-∠ade)/2=(180°-70°)/2=55°.

Then dea+ aic=180°Therefore, the four points a, i, c, and e are on the same circle.

dei=∠cai=25°.

The sum of the inner angles of the triangle is equal to 180°, i.e., b+ c+ bac= AEF+ EF+ EAF=180°

ab=ac abc is an isosceles triangle, i.e. b= c and aef=2 b= b+ c

BAC= EAF+BAE=EAF+ AFE, so AFE= BAE

Do point e, a line parallel to ab, and ac and point e

The specific process is a bit troublesome to look at the computer playing process.

AED AFE can be proven using AAA methods

So ad:ed=ae:ef

ed=2 3ab,ad=1 3ac,ab=ac (this is always fine, right?) ）

ae:ef=1/3:2/3=1:2

Ask the landlord not to use the geometry sketchpad, what are you going to do?

You're so good that you can find such a topic.

1. Solution: Let the radius be r

If od is connected, od=r

De bisects oa perpendicularly, so oc=r2

Vertical diameter theorem, dc=ce=1 2ed=3

Pythagorean theorem. dc²+oc²=od²

3+(r/2)²=r²

3/4r²=3

r = 4r = 2 so the radius is 2

2) Because the angular DPA = 45 degrees, DC is perpendicular AB

So the angle d = 45 degrees.

So the angle eof = 90 degrees (the angle eof is the central angle of the circle).

So the s triangle eof=1 2 2 2=2

s shadow = 1 4 2 2-2 = -2

Take, then s shade =

Question 2 (1) proves: ab is the diameter.

So ADB = 90 degrees, i.e. AD is perpendicular BC

Because ab=ac, according to the three-in-one nature of the triangle.

So AD divides the bac equally

So cad= bad

Because OA=OD

So oad= oda

So od parallel ac

Because DF is vertical AC

So df vertical od

So df is the tangent of the circle od.

2) Because e is the midpoint of the arc AD.

So ae=de

Connecting the OE gives the OE a perpendicular bisector AD (perpendicular diameter theorem).

So ae=de

So ead= eda

Because ead= oad

So eda= oad

So de parallel oa

Because od parallel ae

So the quadrilateral oaed is a parallelogram.

Because OA=OD, OE is perpendicular to AD

So the quadrilateral oade is a diamond.

So oa=ae=oe

So bac=60

daf=1/2∠bac=30

Because df is tangent.

So fde= daf=30 degrees.

de=df cos30=2 ( 3 2)=4 3 That is, circle o radius = 4 3

We can know that the angle aod=120 degrees.

So the arc ad=1 3 circumference = 1 3 2 4 3 = 8 3 9

Question 1 Answer:

1) Make three auxiliary lines first, which are AE, EB, and OF

2) According to known data and conditions, in ΔAce, AE2=AC2+CE2=(1 2R) 2+(2 3 2) 2=1 4R 2+3

In δBCE, EB2=CE2+CB2=(2 3 2) 2+(3 2R) 2=3+9 4R2

In δabe, AB2=AE2+EB2=1 4R2+3+3+9 4R2=5 2R2+6=(2R)2=4R2

Collation equation, 5 2r 2+6=4r 2 3 2r 2=6 r 2=4 r=2

3) Calculate the area of the shaded part:

The area of the shaded part of the sector corresponding to the area of the eof δoef.

According to the triangular principle in the garden, since dpa=45°, therefore, edf=45°, corresponding, , eof=90°

That is, the sector area corresponding to EOF Circle area 4= r 2 4=2 2 4=

Area of OEF = 1 2 r 2 = 1 2x2 2 = 2

The area of the shaded part = -2=

Q2 Answered:

take it easy!

I'm not here for the score, but I hope my words will help you.

First of all, you have good grades, you are worried about the decline in grades, as long as you can stabilize your grades, you can go to your ideal high school. In the final semester, if you have completed all your classes, you will enter early to fill in the gaps. As the saying goes:

The emaciated camel is bigger than the horse, and there is a gap between people and you, and it is still difficult to catch up with you at once. In the same way, for yourself, how you learned before, and now you are like this, or work a little harder, otherwise you will not be able to bear it, and with your grades, it is impossible not to work hard.

Calm down, don't put too much pressure on yourself, if you don't want to play a ball game in class, there is no need to cancel this for the so-called high school entrance examination.

Remember the combination of work and rest, this is easier said than done, and you must not leave your studies aside for the so-called ease.

As for the learning method, your previous one suits you so well, why change to an unknown one that doesn't know if it will suit you. I think your teacher or parent should advise you to take a break early or something, instead of saying go read a book, right?

(1) The connection DE and DF can be verified by constructing similar triangles, and the relationship between AE, AB and af, AC and AD can be obtained by proving that the triangles AED, ADB and triangles AFD and ADC are similar, and the proportional relationship can be obtained through the intermediate value of AD

2) is still true, because it can be proved that the two triangles in (1) are similar, so that the conclusion in (1) can be reached, and in the process of translating up on BC, the conditions for the similarity of the two triangles (a common angle, a set of right angles) have not changed, so they are still similar, so the conclusion in (1) is still true

Ad is the diameter of the circle O, aed=90°

and bc cut circle o to the point d, ad bc, adb = 90°

In RT AED and RT ADB, EAD = DAB, RT AED RT ADB

aead=adab, i.e., ae ab=ad2

In the same way, if we connect DF, we can prove RT AFD RT ADC, AF AC=AD2

ae•ab=af •ac．

2) ae ab = af ac is still true

Ad is the diameter of the circle O, aed=90°

d ab= ead

rt△ad′b∽rt△aed

abad=ad′ae

ae•ab=ad′•ad

In the same way, af ac=ad ad

ae•ab=af•ac

1. Proof: AEF is similar to ACB (angular angle, chord tangent angle = circumferential angle corresponding to the clamped arc AEF = ACB).

2. Ideas: The proof method of this question is related to the previous question, so think about how the second question can be transformed into the first question. Then I will pass the point d to make the tangent of the circle l, and ae and af intersect the straight line l at the points m and n respectively.

ANM, then, is an intermediary, which is similar to AEF (the same as the first question) and ACB (BC is parallel to MN). Certification.

This kind of question, you grasp the general idea, that is, the last two questions continue the proof method of the first question, sometimes, the letters you write are consistent with the first question, you need to experience more. The more you see, the more you know, how to transform.

Thank me for helping you.

1. Proof of connection: EF, AEF = ACB (chord tangential angle = circumferential angle corresponding to the clamped arc) and EAF = CAB (common angle).

aef∽△acb

2. I don't want to write, let's look at that bear baby, oh!

Do og vertical ef in g

OC = Odae Parallel OG Parallel BF

Parallel lines are divided into segments proportional.

oe=ofce=df

Same as above, unchanged.

Upstairs, ef ab, how to get ce=df.