As shown in the figure, point h is the vertical center of △ ABC, the circumscribed circle ⊙ O2 of ⊙ O1 and ⊙ BCH with ab as diameter intersect at point D, and extend ad to intersect ch at point P, Verification: point P is the midpoint of Ch

As shown in the figure, point h is the vertical center of △ ABC, the circumscribed circle ⊙ O2 of ⊙ O1 and ⊙ BCH with ab as diameter intersect at point D, and extend ad to intersect ch at point P, Verification: point P is the midpoint of Ch

It is proved that: as shown in the figure, extend AP to ⊙ O2 at point Q,
Connect ah, BD, QB, QC, QH
Because AB is the diameter of ⊙ O1,
So ∠ ADB = ∠ bdq = 90 °. (5 points)
So BQ is the diameter of ⊙ O2
So CQ ⊥ BC, BH ⊥ HQ. (10 points)
Because the point h is perpendicular to △ ABC, ah ⊥ BC, BH ⊥ AC
So ah ‖ CQ, AC ‖ HQ,
Acqh is a parallelogram. (15 points)
So point P is the midpoint of Ch. (20 points)

As shown in the figure, point h is the vertical center of △ ABC, the circumscribed circle ⊙ O2 of ⊙ O1 and ⊙ BCH with ab as diameter intersect at point D, and extend ad to intersect ch at point P, Verification: point P is the midpoint of Ch

It is proved that: as shown in the figure, extending AP intersection ⊙ O2 at point Q, connecting ah, BD, QB, QC, QH. Because AB is the diameter of ⊙ O1, so ᦼ ADB = ∠ bdq = 90 °. (5 points), BQ is the diameter of ⊙ O2. So CQ ⊥ BC, BH ⊥ HQ. (10 points) and because point h is the vertical center of ⊥ ABC, ah ⊥ BC, BH ⊥ AC

As shown in the figure. The high ad of △ ABC intersects with be at h, and BH = AC. it is proved that: ∠ BCH = ∠ ABC

It is proved that the high ad of ABC intersects with be at h,
∴∠ADB=∠AEB=90°，
∠DBH=90°-∠DHB，∠HAE=90°-∠AHE，
∵∠DHB=∠AHE，
∴∠DBH=∠HAE，
∵BH=AC，
∴△ADC≌△BDH，
∴AD=BD，CD=HD，
∴∠BCH=∠ABD=45°．

As shown in the figure, △ ABC is an isosceles right triangle with right angle side length a, right angle side AB is the diameter of semicircle O1, and semicircle O2 passes through point C and is tangent to semicircle O1, then the area of shadow in the figure is______ .

Let the radius of the semicircle O2 be X. according to the Pythagorean theorem, we can get (A2) 2 + (a − x) 2 = (A2 + x) 2. The solution is: x = A3.

As shown in the figure, the triangle ABC is an isosceles right triangle, D is the midpoint of the circle, BC is the diameter of the semicircle, ab = BC = 10 cm, calculate the area of the shadow part

Connecting BD, OD and OA, because do ⊥ BC, ab ⊥ BC, so do ∥ AB, then s ⊥ AOD = s ⊥ BOD, while the shadow area = s △ AOB + s sector bod-s △ AOD, = s △ AOB + s sector bod-s △ BOD, = 12 × 10 × 10 ﹥ 2 + 14 × π × (102) 2-12 × 102 × 102, = 25 + 19.625-12.5, = 32.125

The extension lines of circle O1 and circle O2 intersect at points c, D, o2o1 and circle O1 intersect at a, AC, ad, and the extension lines of circle O2 intersect with E, F, respectively Verification CD ‖ ef CE=DF

AO2 vertical bisector CD,
Then ad = AC, triangle ACD is isosceles triangle
According to the intersecting string theorem: AD / AC = AE / AF = 1,
Then AE = AF,
DF=AF-AD,
CE = ae-ac, so CE = DF
Angle ACD = angle ADC,
Angle AEF + angle CDF = 180, angle ADC + angle CDF = 180
Then AEF = angle ADC, similarly, angle DFE = angle ACD,
If angle ACD = angle ADC, then angle AEF = angle DFE = angle ADC = angle ACD,
The same position angle is equal, so
CD‖EF

It is known that the radius of circle O1 is 1.5 times of the radius of circle O2, and the length of arc AMB is exactly equal to the circumference of circle O2 Find the degree of the center angle of arc AMB

Circumference = diameter x 3.14
Arc = circle x center angle / 360
The diameter of the two circles is 1:1.5 = 2 / 3:1
Center angle = 360 * 2 / 3 = 240 (degrees)

Circle O1 and circle O2 are circumscribed at point a, and one of the outer tangent lines of the two circles is tangent to point B. If AB is parallel to the other common tangent of the two circles Find the ratio of radius of two circles

The ratio of the radii of the two circles is 1:3
You can push it down first. The angle between the two tangent lines is 60 degrees. Try it

Given that the radii of ⊙ O1 and ⊙ O2 are 3cm and 4cm respectively, and the length of one outer tangent of two circles is 4 √ 3, then the position relationship between the two circles is obtained

The square of center distance = (R1-R2) ^ 2 + the square of tangent line length = 1 + 48 = 49
Therefore, if the center distance is 7 = R 1 + R 2, then the position relationship between the two circles is circumscribed

As shown in the figure ⊙ O1 intersects ⊙ O2, P is a point on ⊙ O1. If the tangent of two circles is made through point P, then the number of tangent lines is O1 is the big circle, O2 is the small circle A 1,2 B 1,3 C 1,2,3 D 1,2,3,4 Which one It's better to analyze it This is today's homework I'm waiting If the radius of two circles is r r, the distance between the centers of two circles is D, and the square of R + the square of D - r = 2rd (R is greater than R), then the position relationship between the two circles is

Choose C. point P on the circle O1, because there is only one tangent line for the circle through the circle, and there are only two tangents for a known circle outside the circle, so at most three tangents can be made