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As shown in the figure, ad vertical BC BD = DC point C on the vertical bisector of AE, what is the relationship between the length of AB AC CE and ab + BD and de 1. As shown in the figure, ad vertical BC, BD = DC, point C is on the vertical bisector of AE, what is the relationship between the length of AB AC CE and ab + BD and de?
Do not look at the title, a look at the problem will know, if not equal, or > or
As shown in the figure, Po is the secant of circle 0, intersecting circle O at A.B, PD tangent circle O to D, AC is a chord of circle O, and PC = PD As shown in the figure, Po is the secant of circle 0, intersecting circle O at A.B, PD tangent circle O to D, AC is a chord of circle O, and PC = PD. (1) verification: PC is tangent of circle O; (2) if AC = PD, verification: BP = OA
Link BC
From 1. We know that ∠ OCP = 90 °
∵ AB is the diameter
∴∠ACB＝90°
∴∠OCP＝∠ACB
∵AC＝PD＝PC
∴∠OPC＝∠BAC
∴△OCP≌△BCA
∴OP＝AB＝2OB
∴BP＝OB＝OA
Hope to be satisfied!
As shown in the figure, AB is the chord of circle O, ab = 12, PA tangent circle O to a, Po ⊥ AB to C, Po = 13. Find the length of PA As the title Just a circle, that, that, that
Connect OA. Then OA ⊥ PA
∵PO⊥AB,AB=12
∴AC=6
Easy to prove △ APC ∽ OAC
∴AC²=PC*OC
Let PC = X
Then x (13-x) = 36
The solution is x = 4 or 9
When PC = 4, PA = 2 √ 13
When PC = 9, PA = 3 √ 13
As shown in the figure, in the triangle ABC, the circle O with the diameter of AB intersects BC at point P, PD intersects with D perpendicular to AC, and PD is tangent to circle O (1) AB = AC (2) BC = 6, ab = 4 to find CD Why is p the midpoint of BC
(1) Is it proof
Connect Po
If DP is tangent to a circle, Op ⊥ DP
And DP ⊥ AC
Then AC is parallel to Op
Then ∠ OPD = ∠ C (isometric angle)
And inside the circle
OP=OD
∴∠OPD=∠ODP
Then ∠ ODP = ∠ C
In CAD, ad = AC
(2) Do AF ⊥ CD in F through a
In isosceles triangle CDA, AF bisects CD
Then CF = DF = 3
AC = ad = 4
And △ ACF is similar to △ PCD
So CP: CD = CA: CF = 4:3
Then CD = 9 / 4
Add: why is p the midpoint of BC
Connect op
OP is parallel to AC
Then △ ACD is similar to △ OPD
And the similarity ratio is od: ad = 1:2 (it can be said that OP is the median line)
Then PD:CD=1:2
So PD = CP = cd-pd
As shown in the figure, in △ ABC, D is the midpoint of AB, the intersection of straight lines passing through point D is AC at point E, and the extension line of intersection BC is at point F. verification: BF: CF = AE: EC ditto
Proof: (draw a picture myself, Khan. I drew it according to your intention.)
Pass through point a as AP ∥ BC, cross DF reverse extension line at P,
∵ D is the midpoint of AB,
∴AD = CD
∵AP‖BF,
∴△APD≌△BFD
∴AP = BF
And AP ‖ BF,
∴△APE∽△CFE,
∴AP/CF = AE/EC
Therefore, BF / CF = AE / EC
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 DF ∥ be is made by crossing point D, crossing AC to F, ᙽ ad is the center line on the edge of BC, that is BD = CD,
This is the inverse use of the median line of a triangle. ad is the center line on the side of BC, that is, BD = CD. We can know that DF is the median line of the triangle BCE, because it is parallel and one point is the midpoint
As shown in the figure, △ ABC is an equilateral triangle, ad ⊥ BC is in D, CE ∥ AB, and AE ⊥ EC proves that AE = ad
∵ △ ABC is an equilateral triangle, ad ⊥ BC is in D
∴∠ABD=∠ACD=60°,∠ADB=∠ADC=90°
And ∵ CE ∵ ab
∴∠ACE=∠ABD=60°=∠ACD
And AE ⊥ EC
∴∠AEC=90°=∠ADC
AC=AC
To sum up, △ ADC ≌ △ AEC (according to AAS)
∴AE=AD
As shown in the figure, in the triangle ABC, if De is parallel to BC, ad = 3, AE = 2, BD = 4, try to find the value of AE than AC, and the length of AC and EC
∵DE‖BC
∴∠ADE=∠ABC,∠AED=∠ACB
∴⊿ADE∽⊿ACB
∴AD∶AB=AE∶AC
∴3∶(3+4)=2∶AC
∴AC=14/3
∴EC=14/3-2=8/3
As shown in the figure, △ ABC, de / / BC, AD / BD = AE / EC, EC / AC = BD / ab
prove:
Let AD / BD = AE / EC = K,
Then ad = KBD, AE = KEC,
Then AB = AD + BD = (K + 1) BD, AC = AE + EC = (K + 1) EC,
∴EC/AC=1/(k+1),BD/AB=1/(k+1),
∴EC/AC=BD/AB
Get the certificate
In the triangle ABC, D is on AB, e is on AC, De is parallel to BC, ad = EC, DB = 1, AE = 4, BC = 5. Find the length of de I only started to learn similar triangles recently. I hope I can understand thx
Let ad = EC = X
ABC is parallel, so BC is similar
So ad / AB = AE / AC = de / BC
Then there is the following equivalent X / (x + 1) = 4 / (x + 4) = de / 5
X / (x + 1) = 4 / (x + 4) gives x ^ 2 = 4. Because X represents length, x > 0 x = 2
DE/5 = 4/(2+4) = 2/3
DE = 10/3
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