A geometry problem about circles in the third year of junior high school, and a geometry problem abo

Let the radius of the bend be r, od=r-cd=r-45

oc ab, oc also bisects ab (theorem).

ad=1/2ab=150

According to the Pythagorean theorem, the column equation is obtained:

r²=(r-45)²+150²

Solution: r=

Let the radius be r, then in the RT triangle AOD, (r-45) 2+150 2=r 2

Solve r, you can do it.

Arc AC = Arc BC

So aoc= cob

aoc=∠cob=120/2=60

ao=oc (both are the radius of a circle).

OAC = OCA (isosceles triangles with equal base angles).

oac+∠oca+∠aoc=180

2∠oac+60=180

oac=60

oca=∠oac=60

So the triangle AOC is an equilateral triangle.

oa=oc=ac

In the same way, the triangle OCB is an equilateral triangle.

oc=bc=ob

So, oa=ac=cb=bo

OACB is diamond-shaped.

Proof: Connect the OC

c is the midpoint of the arc ab, aob=120°

aoc=60°

oa=oc AOC is an equilateral triangle.

oa=ac, the same way, ob=bc

oa=oboa=ac=bc=ob

The quadrilateral oacb is a diamond shape.

Proof: Because :c is the midpoint of the arc ab.

So: aoc= boc

And because: aob=120 degrees.

So: AOC = BOC = 60 degrees.

So the triangle AOC is an equilateral triangle.

So: ao=ac=co

The same can be said: bo=bc=co

Then: a0=bo=bc=ac

So: the quadrilateral oacb is a diamond.

Proof : A , b are two points on o, and, aob=120 degrees, c is the midpoint of the arc ab.

aoc=∠boc=60°

and oa, ob, oc are the radii of the circle o.

oa=ob=oc.

AOC and BOC are equilateral

oa=ac,ob=bc

and oa=ob

oa=ac=ob=bc

The quadrilateral OACB is diamond-shaped.

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

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°

The first question, connect to CO, isosceles, to the top angle, don't need to explain more, right?

Alas, let's write it anyway - fcb=fbc, hcb=hgc=done. The second question is simpler, hmd=heb=dab=eab=emb=mhe+meh

The secant theorem, the circumferential angle of the same arc, the perpendicular theorem, the circumferential angle of the same arc, and the outer angle are used respectively.

Got it......