A proof problem about circles, math masters come soon...

Connect OA, OE, of AE, OA, OE is the bisector of Bao and FEC.

ab=af=1，fe=ec

Let ec=x,de +ad =ae, (1-x) +1 =(1+x) x=, de=,ae=1+:ae=3:5

Take a look at it.,There's no scratch paper around.,Don't blame me for a mistake.,O( o.

EF ec are all tangents of the circle O.

So let ef = ce = x , then: de = 1- x , the same way: ab = af = 1 , af = 1 + x in rt ade:

1 +1-x) = (1+x) solution: x = 1 2

So de = 1 2 , ae = 3 2de: ae = 1 :3

Connect of perpendicular ae, i.e., because ab is also tangent semicircle, i.e., angle abo=90°, angle dfo=90°, according to the law of quadrilaterals, 360°, and because the angle corresponding to the arc is half of the central angle of the circle, the angle baf=60°, i.e., the angle ead=30°, so according to the Pythagorean law, de:ae=1:2

Connecting od, oe, d, e are the midpoints of the arc ab and the arc ac, so od ab, oe ac, [perpendicular diameter theorem].

od=oe, ode= oed,dfb=90°- ode=90°- oed= egc, afg= dfb, agf= egc,[top] so afg= agf,af=ag.

∠bod = aoc + 2∠dpb

Proof: Connect AD

So BOD = 2 bad (the central angle of the same arc is twice the circumferential angle) and the same goes for aoc = 2 adc

And bad = adc + dpb (the outer angle is equal to the sum of the two inward angles that are not adjacent), so bod = 2 bad = 2 adc + 2 dpb = aoc + 2 dpb is proved.

Proof: Because OP is the bisector of the angle OCD, so the angle DCP=Angle OCP, and because OC=OP, the angle OCP=Angle OPC, so the angle DCP=Angle OPC, so the CD is parallel to the OP, and because the CD is perpendicular AB, so the OP is perpendicular AB, so the arc AP is equal to the arc BP, so PA=PB

Note that o should be the midpoint of the triangle.

The point of O is made to be OE perpendicular to AC, and the vertical foot is E

Because o is the midpoint of the triangle.

So ad=ae

And because ab=ac

So angular dao = angular dac

And because ao=ao

So the triangle ado is similar to aeo

So. od=oe

And because od is the radius.

So OE is also a radius.

So ac is the tangent of the circle o.

Hope it works for you.

Be careful when asking questions in the future.

Key points: According to the Pythagorean theorem, fc 10 can be obtained

If you connect the od, then the od ef is known by the tangent

As ON BC, let the radius be 5x, then fa 10 10x is obviously OCN FCE

So you can get on oc ef fc 4 5

So on 4x

Apparently the quadrilateral oden is rectangular.

So ed on 4x

So FD 8 4X

Because EF cuts the circle o to d

So FD 2 FA*FC

So (8-4x) 2=(10-10x)*10 gives x 3 4

So the radius r 5x 15 4

FYI! jswyc

Proof: Connect AO and BO, connect DO

PA and PB are both tangential lines, AB is tangent point strings, and PO AB is obtained according to the theme.

pa2=pc*pd

In the right angle of PAO, am is the height of the hypotenuse, PA2=PM*POPC*PD=PM*PO

i.e. pc po=pm pd

cpm= opd

cpm∽△opd

pmc=∠pdo……This is very important, and will be used in a moment) In the right-angle pao, am is the height of the hypotenuse, ao2=om*op and oa=od

od2=om*op

i.e. om od=od op

and mod= dop

dom∽△pod

dmo=∠pdo……②

Combine and , to get PMC= DMO

am⊥po∠amp=∠amo=90°

amp-∠pmc=∠amo-∠dmo

amc=∠amd

Proven. Thank you.