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O is the diameter of a perpendicular point on each other (1) Verification: AC bisection ∠ DAB; (2) If CD = 4, ad = 8, try to find the radius of ⊙ o
(1) Connect OC,
∵ CD is tangent,
∴OC⊥CD．
∵AD⊥CD，
∴AD∥CO，
∴∠1=∠4．
∵∠2=∠4，
∴∠1=∠2．
(2) Do OE ⊥ ad, and set the radius as X,
∵CD⊥AD，
∴OE∥CD；
OC ⊥ CD,
∴OC∥AD，
The quadrilateral oedc is a rectangle,
∴OE=CD=4，AE=8-x，
∴42+（8-x）2=x2，
∴x=5．
The extension line of circle O diameter AB and chord CD intersect at point P, e is a point on circle O, AE arc = AC arc, de intersects AB at point F The extension line of circle O diameter AB intersects with the extension line of chord CD at point P, e is a point on circle O, AE arc = AC arc, de intersects AB at point F. verification: pf * Po = PD * PC
prove:
Connect OC and OE
Then ∠ COE = 2 ∠ CDE
∵ arc AC = arc AE
∴∠AOC=∠AOE
∴∠AOC=∠CDE
∴∠COP=∠PDF
∵∠P=∠P
∴△PDF∽△POC
∴PD/PO=PF/PC
∴PF*PO=PD*PC
As shown in Fig. 1, it is known that AB is the diameter of circle O, AB is perpendicular to the chord CD, the perpendicular foot is m, the chord AE intersects with the chord CD and F, ad ^ = AE times AF! Hope to hear back in 30 minutes
1. Auxiliary line: connecting EB and DB
2. In triangle AMF and Abe, af * AE = AB * am because AF / AB = am / AE
3. In triangle AMD and ADB, ad * ad = AB * am because AD / AB = am / ad
4. So ad * ad = af * AE
It is known that the diameters AB and CD of circle O are perpendicular to each other, and the chord AE intersects CD with F. if the radius of circle O is r, it is proved that AE * AF = the square of 2R
prove:
Connect be
∵ AB is the diameter
∴∠E=90°
∴∠E=∠AOF
∵∠A=∠A
∴△AOF∽AEB
∴AF/AB=AO/AE
∴AF*AE=AO*AB=R*2R=2R²
AB is the diameter of O, AC, CF is the chord, the chord CD is vertical, AB, AF, CD intersect the point E, and AE = CE, verification: AC = CF
Connect BC
∵ AB is the diameter
﹤ ACB = 90 ° (the circular angle on the semicircle is a right angle)
∵CD⊥AB
∴∠ADC=∠ACB=90°
∵∠CAD=∠CAB
∴△ACD∽△ACB
∴∠ACD=∠ABC
∵AE=CE
Ψ ace = ∠ CAE, i.e., ∠ ACD = ∠ caf
∴∠ABC=∠CAF
﹤ AC = CF (equal circumference angles and equal chords)
The triangle ABC is inscribed in the circle O and ab = AC, point D is on the circle O, ad is perpendicular to point AB, ad and BC intersect at point E, f is on the extension line of Da and AF = AE 1. Determine the position relationship between BF and circle O and explain
Tangency
In the triangle ABC, AE is vertical AB, AE = AB, AF is vertical AC, AF = AC, ad is vertical BC, vertical foot is D, lengthening Da, crossing EF to m, trial evidence: EM = FM
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In △ ABC, ab = AC, AF ⊥ BC, point D is on the extension line of Ba, point E is on AC, and ad = AE. This is to explore the positional relationship between de and AF and prove your conclusion Δ ABC is isosceles triangle, △ ade is isosceles triangle, ∠ AFC = 90 °
DE‖AF
Find a point E on AC, extend Ba, take ad = AE and connect De
∵ △ ABC is an isosceles triangle (a is the vertex, B is on the left and C is on the right) AF ⊥ BC
∴∠FAC=1/2∠BAC
∵ D on Ba extension line
∴∠DAE=180°-∠BAC
∵ △ ade is an isosceles triangle
∴∠AED=(180°-∠DAE)÷2
=[180°-(180°-∠BAC)] ÷2
=1/2∠BAC
∴∠FAC=∠AED
∴DE‖AF
As shown in Fig. 1, in the equilateral triangle ABC with side length 1, D and E are the points on the sides of AB and AC respectively, ad = AE, f is the midpoint of BC, AF and de intersect at point G, and △ ABF is folded along AF to obtain a-bcf as shown in Fig. 2, where BC= Two 2． (1) Proof: de ‖ plane BCF; (2) Proof: CF ⊥ plane ABF; (3) When ad = 2 When 3, calculate the volume vf-deg of the triangular pyramid f-deg
(1) In AD, the triangle is equilateral
DB＝AE
EC is also true in the folded pyramid a-bcf,
∴DE∥BC．
And ∵ de ⊄ plane BCF, BC ⊂ plane BCF,
ν de ‖ plane BCF
(2) In the equilateral triangle ABC, f is the midpoint of BC, so AF ⊥ BC, that is AF ⊥ CF ①, and BF = CF = 1
2．
∵ in a-bcf, BC =
Two
2，∴BC2=BF2+CF2，∴CF⊥BF②．
And ∵ BF ∩ AF = f, ∩ CF ⊥ plane ABF
(3) From (1) we can know that GE ∥ CF, combined with (2), we can get Ge ⊥ plane DFG
∴VF−DEG＝VE−DFG＝1
3•1
2•DG•FG•GE=1
3•1
2•1
3•(1
3•
Three
2)•1
3＝
Three
324．
As shown in the figure RT △ ABC, AC is perpendicular to BC, ad is bisected ∠ BAC intersects BC at point D, and De is perpendicular to ad intersecting AB at point E M is the midpoint of AE, BF is perpendicular to BC, and the extension of CM is at point F 1. AC / BF = CD / BD 2. If BD = 4, CD = 3, the value of be * AC is proved
As shown in the figure, a parallel line passing through point E as BF intersects CF at point G and connects DM
Because AC ⊥ BC, BF ⊥ BC
So, AC / / BF
However, eg / / BF
So, AC / / EG
M is known as the midpoint of AE
So, △ AMC ≌ △ EMG
So, AC = eg
Then, AC / BF = eg / BF
Eg / BF = mg / MF
Because mg = cm
Therefore, AC / BF = eg / NF = mg / MF = cm / MF
Known ad ⊥ De, M is the midpoint of AE
So, am = MD
Therefore, ∠ 2 = ∠ 3
It is known that ∠ 1 = ∠ 2
Therefore, ∠ 1 = ∠ 3
So MD / / AC
So MD / / BF
So cm / MF = CD / BD
Therefore, AC / BF = DC / BD
It is known that ∠ 1 = ∠ 2, ∠ ACD = ∠ ade = 90 °
Therefore, △ ACD ∽ ade
Therefore, ∠ ADC = ∠ AED, and AC / CD = ad / De
Then, ∠ 2 + ∠ abd = ∠ 4 + ∠ abd
Therefore, ∠ 2 = ∠ 4
Therefore, △ ADB ∽ DEB
Therefore, AD / de = BD / be
So, AC / CD = BD / be
Therefore, ac * be = BD * CD = 12
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