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When is it necessary to make auxiliary lines and when is it not necessary to make auxiliary lines?
In general, if you can't solve it directly, you need to use auxiliary lines. Pay attention to whether there are any independent points or lines. This is to prompt you to do auxiliary lines, or there is no direct connection between known and unknown. In fact, there are many kinds of solutions in solid geometry, some need auxiliary lines, but some don't need them
It is known that AB is the diameter of circle O, ad, BC and CD are the tangent of circle O, and the tangent points are a.b.e. it is proved that (1) OC ⊥ OD, (2) ofeg is a rectangle
It's urgent before 24 o'clock today
Boss, you haven't written all the questions. The second question is not easy to prove for you. The first question can be answered by
If CD is parallel and equal to AB, then CE = OE = radius r, OE ⊥ CD, then RT triangle CEO is isosceles right triangle, then angle EOC = angle OCE = 45 °, similarly, angle EOD = 45 °, so angle cod = angle COE + angle EOD = 90 °, that is OC ⊥ OD
Circle O is the circumscribed circle of △ ABC with ab as the diameter. Point D is the midpoint of inferior arc BC. It connects AD and extends. It intersects the tangent passing through point C at point P, OD intersects BC at point E. when AC = 6 and ab = 10, the length of tangent PC is calculated
∵ a, B, D, C are in the same circle, arc CD = arc BD, ∵ CAD = bad = BAC / 2. ∵ AB is the diameter of ⊙ o, ∵ AC ⊥ BC. ∵ cos ∵ BAC = AC / AB = 6 / 10 = 3 / 5, cos ∵ bad = ad / AB = ad / 10
In the circle O with radius of 2.5, there are fixed point C and moving point P on different sides of diameter ab. it is known that BC ∶ CA = 4 ∶ 3, point P moves on arc AB and passes through the arc
We've done this before`
(1) If CP ⊥ AB is symmetric to point c about AB, let d be the perpendicular foot
∵ AB is the diameter of ⊙ o, ∵ ACB = 900
∴AB=5,AC:CA=4:3,∴BC=4,AC=3.
And ∵ AC &; BC = AB &; CD ∵
In RT △ ACB and RT △ PCQ,
∠ACB＝∠PCQ=900,∠CAB＝∠CPQ,
Rt△ACB∽Rt△PCQ
∴
(2) When point P moves to the midpoint of arc AB, it passes through point B and makes be ⊥ PC at point E (as shown in the figure)
∵ P is the midpoint of arc AB, ∵
And ∠ CPB = ∠ cab ℅ CPB = Tan ∠ cab=
Follow
From (L), we get that,
(3) When the point P moves on the arc AB, there is a constant
Therefore, when PC is maximum, CQ is maximum
When PC passes through the center O, that is, the maximum value of PC is 5, the maximum value of CQ is
It is known that AB is the diameter of ⊙ o, P is the point on AB, C and D are the two points on the circle, and they are located on the same side of ab. angle CPA = angle DPB. It is proved that cdpo has four points on the same circle
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The distance from the center of the circle to the points on the circle is equal to prove that the four points are in the same circle
The center of the circle e is the center of the center of the circle
Now we just need to prove EP = EO or EC = ed
It is proved that EP = EO or EC = ED can be done by proving the congruence of triangles, but auxiliary lines are still needed
As shown in the figure, AB is the diameter of semicircle o, point P is on the extension line of AB, PC cuts semicircle o to point C and connects AC. if ∠ CPA = 20 °, calculate the degree of ∠ a
Connect OC, ∵ PC to cut semicircle o at point C, ∵ PC ⊥ OC, namely ∠ PCO = 90 °, ∵ CPA = 20 °, ∵ POC = 70 °, ∵ OA = OC, ∵ a = ∠ OCA = 35 °
As shown in the figure, it is known that PA is the tangent of circle O, a is the tangent point, PBC is the secant, D is the midpoint of CB, e is the midpoint of OP, try to judge: △ AED is the tangent point
Isosceles right triangle
Even OA, OD
⊙ D is the midpoint of BC, ⊥ OD ⊥ PC
∵ PA is tangent, ∵ OA ⊥ pa
∵ e is the midpoint of OP, ∵ EA = EO = EP = ed
A.o.d.p is concentric and E is the center of the circle
∴∠AED=2∠APC=90
The AED is an isosceles right triangle
PA is the tangent of circle O, PBC is the secant of circle O ∠ APC = 45 ° D is the midpoint of CB to judge why △ AED is a special triangle
E is on the OP and outside the circle, connecting ad, De, AE, Op. please draw a picture by yourself,
I use the Geometer's Sketchpad to draw figures. The actual length is not a special triangle
It is known that AB is the diameter of ⊙ o, point P is a moving point on the extension line of AB, the tangent of ⊙ o is made through P, the tangent point is C, the bisector of ⊙ APC intersects AC at point D, then ⊙ CDP is equal to ()
A. 30°B. 60°C. 45°D. 50°
As shown in the figure, connect OC, ∵ OC = OA, PD bisects ∵ APC, ∵ cpd = ∵ DPA, ∵ a = ∵ ACO, ∵ PC is the tangent line of ⊙ o, ∵ OC ⊥ PC, ∵ CPO + ∵ cop = 90 °, ∵ cpd + ∵ DPA + (∵ a + ∵ ACO) = 90 ° and ∵ DPA + ∵ a = 45 ° respectively, that is ∵ CDP = 45 °. Therefore, select C
As shown in the figure, Pt is the tangent line of ⊙ o, the tangent point is t, M is the inner point of ⊙ o, PM and the extension line of PM intersect ⊙ o at B, C, BM = BP = 2, Pt = 25, OM = 3, then the radius of ⊙ O is______ ．
⊙ PT is the tangent line of ⊙ O. according to the cutting line theorem, pT2 = Pb ⊙ PC; ∵ Pt = 25, BP = 2; ∵ PC = pt2 △ PC = 10; ∵ BC = 8, CM = 6; through O and m, make the diameter of ⊙ o, intersect ⊙ o with E and f; let the radius of ⊙ o be r, then EM = R + 3, MF = R-3; according to the intersecting chord theorem, we get: (R + 3) (R-3) = BM · MC; r2-9 = 2 × 6, that is r = 21. Therefore, the radius of ⊙ o is 21
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