As shown in the figure, it is known that PA is the tangent of circle O, a is the tangent point, AB is the diameter of circle O, and the chord BC is parallel to OP and intersects circle O at point C. It is proved that PC is the tangent of circle o

As shown in the figure, it is known that PA is the tangent of circle O, a is the tangent point, AB is the diameter of circle O, and the chord BC is parallel to OP and intersects circle O at point C. It is proved that PC is the tangent of circle o

prove:
Connect to OC
∵OB=OC
∴∠OBC=∠OCB
∵PO∥BC
∴∠AOP=∠OBC,∠COP=∠OCB
∴∠AOP=∠COP
∵PO=PO,OC=OA
∴△OAP≌△OCP
∴∠OAP=∠OCP
∵ is the tangent, AB is the diameter
∴∠PAO=90°
∴∠OCP=90°
Ψ PC is the tangent of circle o

As shown in the figure, rectangle ABCD, ab = 3, ad = 4, take ad as the diameter of semicircle, m as the upper moving point of BC, can coincide with B and C, am intersects semicircle in N, let am = x, DN = y, find out the functional relationship of Y with independent variable x, and find out the value range of independent variable x

∵ ad is the diameter, ∵ B = ∵ and = 90 °, ∵ AMB = ∵ Dan, ∵ ABM ∵ DNA, ∵ abdn = amda, ∵ 3Y = x4, that is, y = 12x, when m is at point C, X is the maximum, which is 5; when m is at point B, X is the minimum, which is 3; the value range of ∵ x is 3 ≤ x ≤ 5

As shown in the figure, rectangle ABCD, ab = 3, ad = 4, take ad as the diameter of semicircle, m as the upper moving point of BC, can coincide with B and C, am intersects semicircle in N, let am = x, DN = y, find out the functional relationship of Y with independent variable x, and find out the value range of independent variable x

∵ ad is the diameter, ∵ B = ∵ and = 90 °, ∵ AMB = ∵ Dan, ∵ ABM ∵ DNA, ∵ abdn = amda, ∵ 3Y = x4, that is, y = 12x, when m is at point C, X is the maximum, which is 5; when m is at point B, X is the minimum, which is 3; the value range of ∵ x is 3 ≤ x ≤ 5

As shown in the figure, in the right angle trapezoid ABCD, AD / / BC, AB is vertical BC, with ab as the diameter, ab = BC = 1, with ab as the radius, make a semicircle o, cut CD to e, connect OE, and extend the intersection
The extension of ad is in F
(1) Q: can the angle BOE be 120 degrees and give a brief explanation
(2) It is proved that triangle AOF is similar to triangle EDF, and the similarity ratio is 2
(3) Finding the length of DF
OK, I'll add points

Because the network is not easy to send pictures to you, the auxiliary line is connected to AC od OC, the extension line of fo CB is connected to AC at H point (1), because AB = BC, so ∠ ACB = 45 ° because OE ⊥ CD (the line connecting the center of a circle and the tangent point must be perpendicular to the tangent line) ab ⊥ BC, so ∠ EOB + ∠ ECB = 180 °. If ∠ BOE = 12

As shown in the figure, there is a point C on the semicircle o with the diameter of ab. the tangent line of the semicircle passing through point a intersects the extension line of BC at point D. (1) prove: △ ADC ∽ BDA; (2) the parallel line of AC passing through point O intersects BC respectively at two points E and F. if BC = 23, EF = 1, calculate the length of AC

(1) It is proved that: ∵ AB is the diameter, ∵ ACB = 90 °, ∵ ACD = 90 °. ∵ ad is the tangent of semicircle o, ∵ bad = 90 °, ∵ ACD = ∵ bad. And ∵ ADC = ∵ BDA, ∵ ADC ∽ BDA. (2) connect OC, ∵ OE ∥ AC, ∵ OE ⊥ BC, ∵ be = EC = 3. In RT △ OBE, let ob = x, then

As shown in the figure, it is known that C is a point on the semicircle o with ab as the diameter, CH ⊥ AB is at point h, the line AC intersects the tangent passing through point B at point D, e is the midpoint of CH, connecting AE and extending the intersection BD to F, and the line CF intersects the line AB at point g. (1) prove that point F is the midpoint of BD; (2) prove that CG is the tangent of ⊙ o

It is proved that: (1) ch ⊥ AB, DB ⊥ AB, ∧ AEH ∧ AFB, ∧ ace ∧ ADF. (1 point) ∧ EHBF = aeaf = cefd. ∧ he = EC, ∧ BF = FD. (3 points) (2) connecting CB and OC, ∧ AB is the diameter, ∧ ACB = 90 °, ∧ f is the midpoint of BD, CF = DF = BF, ∧ BCF = cab = CBF = 90 ° -

As shown in the figure, AB is the diameter of semicircle o, Ao is the diameter of semicircle m, C is the midpoint of ob, passing through point C is the tangent of semicircle m, intersecting semicircle m at point D, extending ad intersecting circle O at point D

The distance from point a (4, - 3) to the x-axis is___ And the distance to the y-axis is___ The distance to the origin is

The distance from the X axis is the absolute value of the ordinate of point a, which is 3
The distance from the y-axis is the absolute value of the abscissa of point a, which is 4
The distance from the origin is to solve the right triangle. According to the Pythagorean theorem, the square of 3 + the square of 4 =
The Pythagorean number is 3,4,5, so the distance from point a to the origin is 5

Given the distance from point a (- 3, - 4) to the x-axis is______ And the distance to the y-axis is______ The distance to the origin is______ ．

The distance from point a (- 3, - 4) to X axis is 4, the distance to y axis is 3, and the distance to origin is 32 + 42 = 5

The distance from point a (- 6,8) to X axis is, the distance to y axis is, and the distance to origin is
The first two will. How to calculate the distance to the origin?

8. 6; 10 = radical (8 ^ 2 + 6 ^ 2). Pythagorean theorem