In the third year of junior high school, I will not urgently ask for help from well wishers

1. Tangent. Take oa as the radius, so a is on the circle.

Tangent again as the angle OAC is 90 degrees.

2. Angle CAD angle CDA

So cd ca x

In a right-angled triangle OAC.

The square of the oa, the square of the ac, the square of the od.

Solve x 12

So AC 12

ap+pb=2r, there is pb=(2 3)r, ap=(4 3)r connected od, which can be obtained by the perpendicular diameter theorem, od=2 2 cm is used by the pythagorean theorem in the triangle opd, r2=op2+dp2op=r-pb=(1 3)r

The radius of the solution is r=3cm

ab=6cm

Connect the OB and OC

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

ob=oc=oa

oba=∠oab,∠oac=∠oca

Again, oab+ oac= bac

Li Namin Slow circle OAB+ OAC+ OBA+ OCA=180°, that is, 2 OAB+2 OCA=180°

oba+∠oca=90°

Ad BC again

Which ADC = 90°

and dac=30°

dca=180°-90°-30°=60°∠oba=90°-60°=30°

bae=∠oba=30°

Connect BO and CO to get ao Bo CO

Set angle 1 angle bao round world which angle abo, angle 2 angle cao angle aco get angle 1 + angle 2 angle bao + angle cao angle a

Angle B Angle 1 + Angle 2 Angle B Angle Return Key ABO+Angle ACO Angle CBO+ Angle ACO Angle BCO+ Angle ACO Angle C

Angle C 90 Angle DAC 60

Three-style orange code joint, get.

Angle bae angle 1 30 degrees.

In the right-angled trapezoidal ABCD, AD is parallel to BC, B=90°, and AD+BC=CD, and AB is used as the diameter to make a circle O, and it is verified that CD is tangent to the circle 0.

Do OE perpendicular to CD Nexus OD OC

Trapezoidal ABCD surface surface product = triangle OAD+OBC+ODC1 2*AO*AD+1 2*BO*BC+1 2*OE*CD1 2*AO*(AD+BC)+1 2*OE*CD1 2*(AO+EO)*CD

Trapezoidal ABCD area = 1 2*AB*(AD+BC) =1 2*2AO*CD = AO*CD

So 1 2*(ao+eo)*cd = ao*cdao+eo=2ao

Eo=AO, then the distance from the dot O to CD is Oe, and Cd is tangent to the circle O at E

Let's talk about question 8 Draw a rectangle or square in the garden, and then draw a diagonal line inside the rectangle, and the place where the two diagonals coincide is the center of the circle.

Solution: Even om crosses ab in f, and passes o as od perpendicular mn to d, so od bisects mn, in the right triangle odm, om=4, md=mn 2=2 3, so amd=30°, and m is the midpoint of the arc ab, so om ab, so in the right triangle cmf, acm=90- omc=60°