![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
It is known that AC and BD are two mutually perpendicular chords of the circle O: x2 + y2 = 4, and the perpendicular foot is m (1, 2) The maximum area of the quadrilateral ABCD is () A. 4 B. 4 Two C. 5 D. 5 Two
If the distance between O and AC and BD is D1 and D2 respectively, then D12 + D22 = om2 = 3
The area of quadrilateral ABCD is s = 1
2AC•BD=1
2•2
4−d12•2
4−d22=2
4−d12•
4−d22
If and only if D12 = D22, the equal sign is taken,
Therefore, C
As shown in the figure, it is known that in ▱ ABCD, the parallel line Mn of AC intersects Da, the extension line of DC intersects with m, N, crosses AB, BC at P, Q, and proves that QM = NP
It is proved that: ∵ quadrilateral ABCD is a parallelogram
∴MD∥BC，AB∥ND，
∵MN∥AC，
∴MQ∥AC，AM∥QC，PN∥AC，AP∥CN，
The quadrilateral amqc and quadrilateral APNC are parallelograms,
∴MQ=AC，PN=AC，
∴QM=NP．
As shown in the figure, it is known that in ▱ ABCD, the parallel line Mn of AC intersects Da, the extension line of DC intersects with m, N, crosses AB, BC at P, Q, and proves that QM = NP
It is proved that: ∵ quadrilateral ABCD is a parallelogram
∴MD∥BC，AB∥ND，
∵MN∥AC，
∴MQ∥AC，AM∥QC，PN∥AC，AP∥CN，
The quadrilateral amqc and quadrilateral APNC are parallelograms,
∴MQ=AC，PN=AC，
∴QM=NP．
As shown in the figure: in the parallelogram ABCD, the straight line Mn parallel to the diagonal AC intersects DA and DC respectively at points m, N, Ba, BC at points P and Q, and proves MP Proof MP = NQ
prove:
∵ parallelogram ABCD
∴AD∥BC,AB∥CD
∴∠MAB＝∠B,∠M＝∠CQN,∠BCN＝∠B
∴∠MAB＝∠BCN
∵MN∥AC
The parallelogram amqc
∴AM＝CQ
∴△AMP≌△CQN （ASA）
∴MP＝NQ
The math group answered your question,
As shown in the figure, in the parallelogram ABCD, Mn ∥ AC, Mn intersect the extension lines of Da and DC at points m and N, and intersect AB and BC at points P and Q, Mn = 5, and find PN
From Mn ∥ AC, ad ∥ BC, the quadrilateral amqc is a parallelogram, so MQ = AC;
According to Mn ∥ AC, ab ∥ CD, the quadrilateral APNC is a parallelogram, so PN = AC;
Therefore, PN = MQ = 5
As shown in the figure, the angle between the diameter ab of circle O and the chord AC is 30 ° and the extension line of AB passing through the tangent line of point C is at D. if DC = 5, then the radius of circle O is
Because AB is the diameter of circle O, so ∠ ACB = 90 ° and ∠ BAC = 30 ° so ∠ CBA = 60 ° so OC = od = CD, ∠ cod = ∠ CBA = 60 ° OC is the radius of circle O, and CD cuts o to C, so there is OC ⊥ CD, ≌ OCD = 90 ° = ∠ ACB, so △ OCD ≌ BCA, AC = DC = 5, the radius of circle O is r = OC = AC /
As shown in the figure, it is known that AB is the diameter of circle O, point D is on the extension line of AB, and AC = CD, point C is on circle O, and angle cab = 30 degrees. It is proved that DC is the tangent line of circle o
∵AC=CD
∴∠CAB=∠CDB=30°
Connect OC
∵OA=OC
∴∠CAB=∠OCA=30°
∴∠COD=60°
∴∠OCD=90°
C is on circle o
The circle of O is tangent
As shown in the figure, the diameter ab of ⊙ o is 4; the straight line Mn passing through point B is the tangent line of ⊙ o; D and C are two points on ⊙ o, connecting ad, BD, CD and BC (1) It was proved that ∠ CBN = ∠ CDB; (2) If DC is a bisector of ∠ ADB and ∠ DAB = 15 °, find the length of DC
(1) Prove that: ∵ AB is the diameter of ⊙ o,
∴∠ADB=∠ADC+∠CDB=90°，
∵ Mn cut ⊙ o at point B,
∴∠ABN=∠ABC+∠CBN=90°，
∴∠ADC+∠CDB=∠ABC+∠CBN；
∵∠ADC=∠ABC，
∴∠CBN=∠CDB；
(2) As shown in the figure, connect od and OC, and make OE ⊥ CD at point e through point o;
∵ CD bisection ∵ ADB,
∴∠ADC=∠BDC，
/ / arc AC = arc BC,
∵ AB is the diameter of ⊙ o,
∴∠ADB=90°；
∵ DC is the bisector of ∵ ADB,
∴∠BDC=45°；
∴∠BOC=90°；
And ∵ DAB = 15 °,
∴∠DOB=30°，
∴∠DOC=120°
∵OD=OC，OE⊥CD，
∴∠DOE=60°
∴∠ODE=30°，
∵OD=2，
∴OE=1，DE=
3，
∴CD=2DE=2
3．
As shown in the figure, the diameter ab of ⊙ o is 4; the straight line Mn passing through point B is the tangent line of ⊙ o; D and C are two points on ⊙ o, connecting ad, BD, CD and BC (1) It was proved that ∠ CBN = ∠ CDB; (2) If DC is a bisector of ∠ ADB and ∠ DAB = 15 °, find the length of DC
(1) It is proved that: ∵ AB is the diameter of ⊙ o, ADB = ∠ ADC + ∠ CDB = 90 °, ∵ Mn cut ⊙ o at point B, ABN = ∠ ABC + ∠ CBN = 90 °, ∵ ADC + ∠ CDB = ∠ ABC + ∠ CBN; ? ADC = ∠ ABC, ∵ CBN = ∠ CDB; (2) connect OD, OC, point o as OE ⊥ CD at point E; ? CD
As shown in the figure, point D is a point on the extension line of diameter CA of ⊙ o, and point B is on ⊙ o, and ab = ad = Ao (1) It is proved that BD is tangent of ⊙ o; (2) If point E is a point on inferior arc BC, AE and BC intersect at point F, and the area of △ bef is 8, cos ∠ BFA = 2 3. Find the area of △ ACF
It is proved that: (1) the connection of the Bo is: (1) the connection of the Bo, ? AB = AB = ad ∵ d = ∵ abd ∵ AB = ab ? AB = ab ∵ ab ∵ ab ∵ ab ? ab ∵ ab ∵ ab ∵ ab ∵ ab ? AB = ad ∵ D ∵ abd ? ab ? AB = AB e ab ? ab ? ab ? ab ?
RELATED INFORMATIONS
- 1. It is known that AB is the diameter of ⊙ O. the straight line BC and ⊙ o are tangent to point B. through a, ad ⊙ OC intersects ⊙ o at point D and connects CD Proof: CD is tangent of ⊙ o
- 2. As shown in the figure, AB is the chord of ⊙ o, points c and D are on the chord AB, and OC = OD
- 3. As shown in the figure, it is known that AB is parallel to CD, OA = OD, BC passes through point O, e, f are on the straight line AOD and AE = DF
- 4. As shown in the graph, points D, e and F are the points on the edges BC, Ca and ab of the triangle ABC, De is parallel to Ba, DF is parallel to Ca, and it is proved that ∠ FDE = ∠ a
- 5. As shown in the figure, the radius of ⊙ o is known to be 1, De is the diameter of ⊙ o, the tangent of ⊙ o is ad, C is the midpoint of AD, AE intersects ⊙ o at point B, and the quadrilateral bcoe is a parallelogram (1) Find the length of AD; (2) Is BC tangent to ⊙ o? If so, give proof; if not, give reasons
- 6. 0
- 7. AB is the diameter of the circle, the radius OC ⊥ AB, the chord EF is parallel AB through the midpoint D of OC. It is proved that the angle Abe is equal to 15 ° and the angle CBE is equal to 30 °
- 8. In the hexagon ABCDEF, ∠ a = ∠ D, ∠ B = ∠ e, ∠ C = ∠ F, try to guess the position relationship between AF and CD, and explain the relationship
- 9. Two equal line segments AB and CD coincide in one third. M and N are the midpoint of AB and CD respectively. If Mn = 12cm, find the length of ab
- 10. As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 120 ° and EF is the vertical bisector of ab. EF intersects BC at point F and ab at point E. verification: BF = 1 2FC．
- 11. In the triangle ABC, we know ∠ a = 30 ° and ∠ CBD = 90 ° to find the degree of ∠ BCE
- 12. 0
- 13. As shown in the figure, in △ abd and △ ace, ab = ad, AC = AE, ∠ bad = ∠ CAE, connecting BC and de intersecting at point F, BC and ad intersecting at point g. verification: BC = De
- 14. In the triangle ABC, the points D and E are on BC, the angle CAE is equal to angle B, e is the midpoint of DC, BD is equal to AC, and the ad bisection angle BAE (1) when the angle BAC is equal to 90 degrees In the triangle ABC, points D and E are on BC, the angle CAE is equal to angle B, e is the midpoint of DC, BD is equal to AC, and ad bisects the angle BAE (1) When the angle BAC is equal to 90 degrees, it is proved that BD is equal to AC (2) When the angle BAC is not equal to 90 degrees, is there still BD equal to AC? Explain the reason
- 15. As shown in the figure, in the triangle ABC, ad is the bisector of ∠ BAC, EF bisects ad vertically, intersects ad at e, and the extension line intersects BC at F, then ∠ B = ∠ caf? Because ad bisects angle BAC, angle bad = angle CAD Because EF bisects ad vertically, AF = DF, so angle DAF = angle ADF Because angle DAF = angle DAC + angle CAF, angle ADF = angle B + angle bad, angle DAC + angle CAF = angle B + angle bad Because angle bad = angle CAD, angle B = angle caf Why angle ADF = angle B + angle bad
- 16. As shown in the figure, in △ ABC, D is the midpoint of BC, de ⊥ BC, the bisector of ⊥ BAC is AE at point E, EF ⊥ AB is at point F, eg ⊥ AC intersects AC extension at point g. verification: BF = CG
- 17. As shown in the figure, in △ ABC, D is a point on AB, and ad = AC, AE ⊥ CD, the perpendicular foot is e, and F is the midpoint of CB
- 18. As shown in the figure, in diamond ABCD, AE ⊥ BC, perpendicular foot is e, angle DAE = 2, angle BAE, BD = 15, find the length of AC, ab
- 19. As shown in the figure, both the triangle ABC and the triangle Dec are isosceles right triangles. The angle ACB = angle DCE = 90 ° f is the midpoint of De, h is the midpoint of AE, and G is the midpoint of BD If the △ Dec in Fig. 1 is rotated clockwise by an acute angle around point C to Fig. 2, is the triangle FGH still an isosceles right triangle
- 20. As shown in the figure, D, e and F are points on the three sides of △ ABC, and the areas of CE = BF, △ DCE and △ DBF are equal Verification: ad bisection ∠ BAC