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As shown in the figure of triangle ABC, ad is the center line on BC side, f is a point on ad, and AF: FD = 1:3, extend BF, cross AC to e, and find AE: EC
When DM / be is crossed with AC at m, the △ AEF is similar to △ ADM AE: am = AF: ad = 1:4
Delta CDM is similar to delta CBECM:CE=CD :CB=1 :2 ∴CM=ME ∴AE:EC=1:7
As shown in the figure, in △ ABC, ad is the center line on the BC side, e is on the AC side, and AE: EC = 1:2, be intersects ad at P, then AP: PD is equal to () A. 1：1 B. 1：2 C. 2：3 D. 4：3
Pass through point D as DF ∥ be and AC to F,
The ad is the center line on the side of BC,
That is BD = CD,
∴EF=CF，
∵AE：EC=1：2，
∴AE=EF=FC，
∴AE：EF=1：1，
∴AP：PD=AE：EF=1：1．
Therefore, a
In △ ABC, D is the midpoint of BC, ad is the center line of BC edge, e is the point on AB, connecting EC, AE: be = 1:2, ad and CE intersect at point P, then ad: PD =?
From Menelaus theorem
AE/EB*BC/CD*DP/PA=1
So ad: DP = 1:1
As shown in the figure, in the triangle ABC, D is the midpoint of BC, e is a point on AC, AE: EC = 1:2 be intersection AD and point F So AF: FD = what,
Through D, DG ∥ be is made and AC is given to g,
∵ D is the midpoint of BC,
/ / G is the mid point of CE, i.e. eg = 1 / 2ce,
∵AE：CE=1：2,
∴AE=EG,
EF ‖ DG,
ν f is the mid point of AD,
AF: DF = 1
In △ ABC, ab = 12, E on AC, D on AB, if AE = 6, EC = 4, AD / DB = AE / EC, find ad
∵ AD / DB = AE / EC ᙽ AD / AB = AE / AC ᙽ AD / AB = AE / AC ᙽ AD / 12 = 6 / 6 + 4AD / 12 = 6 / 1010ad = 72ad = 7.2 I help you list the detailed steps and calculation process. For example, mjdodo, his answer is definitely deducted in the exam. If you skip the steps, you can look at your own ideas and think about it again
As shown in the figure, AB is the diameter of ⊙ o, CE ⊥ ad is in E, connected with be, CD＝ CB． (1) It is proved that CE is tangent of ⊙ o; (2) If AE = 6, the radius of ⊙ o is 5, find the value of Tan ᦝ bec
(1) It is proved that OC and BD are connected, and they intersect at point F, as shown in Fig,
A kind of
CD＝
CB，
∴OC⊥BD，FD=FB
∵ AB is the diameter,
∴∠ADB=90°，
∴AE∥OC，
∵CE⊥AD，
∴OC⊥CE，
And ∵ OC is the radius of ⊙ o,
⊙ CE is the tangent line of ⊙ o;
(2) Let ed = x, then ad = 6-x,
∵∠DEC=∠EDC=∠DFC=90°，
The quadrilateral EDFC is rectangular,
∴CF=DE=x，
∴OF=OC-CF=5-x，
∵ of is the median line of ∵ abd,
/ / ad = 2of, i.e. 6-x = 2 (5-x), x = 4,
∴OF=1，DE=4，
In RT △ oBf, BF=
OB2−OF2=2
6，
∴BD=2BF=4
6，
∴tan∠DBE=DE
DB=4
Four
6=
Six
6，
∵EC∥DB，
∴∠DBE=∠BEC，
∴tan∠BEC=
Six
6．
AB is the diameter of ⊙ o, points c and D are two points on the circle, and arc CB = arc CD, CF ⊥ AB at point F and CE ⊥ ad at point E Prove that when de = BF: ∵ arc CB = arc CD ∴CD=BC ∠CAD=∠CAB I just want to know why we can introduce ∠ CAD = ∠ cab
Because the circle angles of the same arc are equal
There is this theorem in the ninth grade mathematics textbook. Don't you see it? Take a closer look~
Maybe it's not in your textbook, but in PEP's books
You can also ask me if you have any questions. I learned this chapter very well
As shown in the figure: AC= CB, D and E are the midpoint of radius OA and ob respectively, Confirmation: CD = CE
Proof: connect OC
In ⊙ o, ⊙ o, ∵
AC=
CB
∴∠AOC=∠BOC，
∵ OA = ob, D and E are the midpoint of radius OA and ob respectively,
∴OD=OE，
∵ OC = OC (common side),
∴△COD≌△COE（SAS），
The corresponding sides of an congruent triangle are equal
As shown in the figure, AB is the diameter of circle O, arc CD is equal to arc CB, CE is perpendicular to AD and connected with be, 1. Verification: CE is circle O, tangent 2
It is proved that: connecting OC, BC, because CE is perpendicular to AD and E, AB is the diameter of circle O, so angle CED = angle ACB = 90 degrees, so angle EAC + angle ECA = angle BAC + angle ABC = 90 degrees, because arc CD = arc CB, so angle EAC = angle BAC, so angle ECA = angle ABC, because angle ABC = angle OCB, so angle ECA = angle OCB, because angle OCB +
As shown in the figure, circle O is the circumscribed circle of quadrilateral ABCD, arc CB = arc CD, CE ⊥ AB at point E, and prove that ab = AD + 2be Don't copy, please
It is proved that: make CF ? ad, make ad, the extension line is f, and connect AC ? CE ? ab ? AEC = ∵ AFC = 90 ? CD = CB ? DAC = ≓ BAE (equal chord angle) and ∵ AC = AC = AC ≌ CF = CE ? RT be ∵ AE =
RELATED INFORMATIONS
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