Junior 3 math problems, geometry problems, about circles, urgency, urgency, urgency

First of all, you set ab=ac=bc=a, and the radius of the circle you set as rabc's area = base multiplied by height divided by 2 ( height (af) = 2 points of the root number 3 times a) can you understand?

The root of the 4th number 3 times the square of a.

Speaking of the area of the shaded part, we can see that the area of the shaded part is actually the sum of two congruent right triangles.

That is, the area ofc = area ocg, so the area of the shaded part is = 2 times the area ofc

Now find the area ofc=base multiplied by the height divided by 2

fc multiplied by of 2

2/2 A multiplied by 2/2 R divided by 2

8/8 of the area of the shadow of AR is = 2 times the area ofc = 4/4 of AR, let's talk about the relationship between a and r af=ao+of

The root of 2 is 3 times a = r + 2 r

So there is r=3 of the root number 3 multiplied by a

Substituting the shaded part area = 4/4 of the AR = 3/4 times the root number 3 multiplied by the square of a.

In summary, it can be seen that the area of one-third of ABC = the area of the shaded part.

1, Connecting OA, OB, then the sum of the areas of OAC, OBC is 2 3 of ABC, and the sum of the areas of OGC, OFC is 1 2 of the sum of the areas of OBC, OAC, and the area of the quadrilateral OGFC is 1 3 of ABC

2. After O is the perpendicular line of BC and AC, the perpendicular foot is F, G, OD, and OE respectively cross BC, AC in H, i, then it can be proved that Ofh Ogi, that is, FH = Gi, CF CG is always equal to BC, AC, and the quadrilateral area is always equal to 1 3 of ABC according to the derivation of 1

Tips: In DOE 360° 90° 2 60° 120°, in connection OA, OB, OC, then AOC BOC 120°; If the sector OAEC is rotated 120° clockwise around the point O, then OE(G) and OC are rotated to OD(F) and OB respectively; s ocg s obf, s quadrilateral ogcf s ocg s ocf s ocf s ocf s obc 1 3s abc.

The circumferential angle theorem of the properties of similar triangles is not a difficult problem and should give you the correct answer.

What are you asking for?

Didn't write it clearly.

1.∵ab//dc

acd=∠bac=35

aod=2 acd=70 (the central angle of the same arc is twice the circumferential angle)2BC third divided arc BC

boc=(1/3)∠aod

aod=138

aed=(1/2)∠aod=69

Question 1: 70°. ab cd , so bac= acd=35° because acd is the circumferential angle of the arc ad, aod is the central angle of the arc ad, so aod=2 acd=35°

Question 2: You didn't finish the 、、、 did you ask for an AED? AED is because AOD=3 BOC=138°, and because AED=1/2 AOD=69°

ab is the diameter of the bar:

Extend cp and intersect o at point f

ab cf arc af = arc ac

Arc AC = Arc CE

Arc AF = Arc CE

ACP= CAD (Equal Circumferential Angles Paired) AD=CD

The ab that seems to be missing in the condition is the diameter.

pca＋∠bac＝90°＝∠abc＋∠bac∴∠pca＝∠abc

Arc AC = Arc CE

ABC CAE (Equal Circumferential Angles Paired) PCA CAE

ad＝cd

Extend the intersection of cp with circle o m because c is the midpoint of arc ae so arc ac = arc ce so angle eac = angle cea because cp is perpendicular ab so cm perpendicular ab according to the diameter perpendicular to the chord arc ac = arc am so angle acp = angle aec so angle cae = angle acp so ad=cd

First of all, EF must be on both sides of the diameter, and it is impossible to do it on one side, right?

Because ECA= FCA, then ACB- ACB= ACB- ACF i.e. ECB= FCB, because ECB= FCB, OE=OF, OC=OC, so the triangle ECO and FCO are congruent and proved.

Answer: Known ECA= FCA, ACB- ACE=ACB- ACFECB= FCB

Then ecf is an isosceles triangle.

ec=fc

Connect to om and pass o as od mn

md=nd=1 Qi Nao2mn=2 3, Shen Fu m+ o=90om=4sino= 3 2

o=60m is the midpoint of the arc AB.

om ab m + high filial pie hood acm = 90

acm=∠o=60