Math problem of grade three (come in! )

CD⊥OA, so ∠ADC=∠CDO=90.

So △ADC∽△CDO ∠COD=∠ACD

Because ∠CAD+∠ACD=90, ∠CAD+∠COD=90.

OC⊥AC。 AB is the tangent of circle o.

(2)① Let D be OD vertical AB, and let D be DH vertical OA in H.

PM⊥OA,EN⊥OA。 So PMNE, quadrilateral PMNE is a right-angled trapezoid.

In RT△AOB, OA=6, OB=8. So AB= 10

PE is a chord, OD⊥PE, so PD=DE. D is the midpoint of PE.

DH⊥OA, so DH∨pm.

DH is the trapezoid midline, and PM+EN=2DH.

OD is the height on the hypotenuse of RT△AOB, so OD=OA×OB/AB=24/5.

According to the projective theorem, OA? =AD×AB. So AD= 18/5.

DH is the height on the hypotenuse of RT△OAD, so DH=AD×OD/OA=72/25.

PM+EN=a+b=2DH= 144/25

② let OM be x, then am = OA-om = 6-X.

Because PM⊥OA, OB⊥OA, pm∨ob.

Simple △AOB∽△APM

AM/AO=PM/OB

(6-X)/6=PM/8

6PM=48-8X

PM=-8-4X/3

S trapezoid PMOB= 1/2×(PM+OB)×OM

= 1/2×(8-4X/3+8)×X

=-2X？ /3+8X

It can be seen that when X=-8/(-2/3×2)=6, the area has a maximum value.

X < 6 because OA=6.

So the smaller x is, the smaller the trapezoidal area is.

When E coincides with A, P is closest to B and X is smallest.

At this time, AP=2AD=36/5.

PB=AB-AP= 14/5

For PQ⊥OB in Q, it is simply △AOB∽△PQB.

PQ/OA=PB/AB

PQ=42/25. So OM=42/25.

Replace X=42/25

S- trapezoidal pmob =-1176/625+8× 42/25 = 7224/625.

Who gave such a sick question?