Given that B and C are two fixed points, the absolute value of BC is equal to 8, and the circumference of triangle ABC is equal to 18, the trajectory equation of vertex A is obtained Why is there only one trajectory equation? Isn't there two cases? It is known that F1 and F2 are two focal points of the ellipse, and the point m satisfying the vector MF1 * MF2 = 0 is always inside the ellipse, then the eccentricity range of the ellipse is Why does the interior of the ellipse include boundaries?

Given that B and C are two fixed points, the absolute value of BC is equal to 8, and the circumference of triangle ABC is equal to 18, the trajectory equation of vertex A is obtained Why is there only one trajectory equation? Isn't there two cases? It is known that F1 and F2 are two focal points of the ellipse, and the point m satisfying the vector MF1 * MF2 = 0 is always inside the ellipse, then the eccentricity range of the ellipse is Why does the interior of the ellipse include boundaries?

(1) Where are the two situations? Where do you say the focus is?
This topic requires you to establish a direct coordinate system
Then the equation is solved
The equation is deterministic
(2) The interior of an ellipse, without boundaries

In the triangle ABC, the absolute value of AB is equal to the root 2, and the absolute value of BC is equal to 2. Find the trajectory equation of point a

Establishing rectangular coordinate system with the midpoint of BC as the origin: B (- 1,0) C (1,0)
Let (a, x)
|AB|=√2|AC|
The square of both sides is: (x + 1) ^ 2 + y ^ 2 = 2 [(x-1) ^ 2 + y ^ 2]
It is concluded that: (x-3) ^ 2 + y ^ 2 = 8

It is known that the three vertices of △ ABC are all on the circle O. ad is the height of △ ABC and AE is the diameter of circle O. it is proved that ab * AC = AE * ad

Proof: link be, CE
⊙ o is the circumcircle of ⊙ ABC,
AE is the diameter of ⊙ o
∴∠ABE=90°=∠ACE
Equal circumference angle: ∠ ABC = ∠ AEC
∠AEB=∠ACB
AD⊥BC
∴Rt△ABD∽Rt△AEC
Rt△ABE∽Rt△ADC
∴AB/AE=AD/AC,AB/AD=AE/AC
∴AB·AC=AD·AE

As shown in the figure, the three vertices of △ ABC are all on ＜ o, AB is the diameter, CD bisects ∠ ACB, ∠ cab = 30 ° and find the length of AB, ad and BD

Are the three vertices of the triangle ABC on the circle O, where AB is the diameter = = > ∠ C = 90 °; = > RT Δ cab ∠ cab = 30 ° = = > BC = 1 / 2 * AB; AC = √ 3 / 2 * ab ν AB = AC / (√ 3 / 2) = 2 √ 3; BC = 1 / 2 * AB = √ 3 CD bisection angle ACB = = > ad: BD = AC: BC =

As shown in the figure, in the triangle ABC, ad is the midline on BC. (3) explore the relationship between the sum of AB and AC in the triangle and the midline ad, and explain the reasons; (4) If AB = 5, AC = 3, then what is the range of ad?

AB＋AC＞2AD
It is proved that: extend ad to point e so that ED = ad,
It is easy to prove that △ ADC ≌ △ EDB,
∴AC=EB,
In △ Abe, from the trilateral relationship, it is concluded that:
AB＋EB＞AE,
That is, AB + AC > 2ad

In the triangle ABC, ad is the center line on the side of BC. Try to guess the quantitative relationship between (AD + BD) and 1 / 2 (AB + AC), and explain the reasons Delta

AD+BD＞AB（1)
AD+cD ＞AC（2）
（1）+（2）
2AD+BD+CD＞AB+AC
2AD+2BD＞AB+AC
AD+BD＞1/2（AB+AC）

As shown in the figure, △ ABC, we know ∠ BAC = 45 ° ad ⊥ BC in D, BD = 2, DC = 3, and find the length of AD Xiao Ping used the knowledge of axisymmetry to turn the figure and solve the problem skillfully Please follow Xiaoping's ideas to explore and answer the following questions: (1) Taking AB and AC as the symmetry axes, the axisymmetric figures of △ abd and △ ACD are drawn. The symmetry points of point D are e and F, and extended EB and FC intersect at point g. it is proved that the quadrilateral aegf is a square; (2) Let ad = x, use Pythagorean theorem to establish the equation model of X, and get the value of X

(1) It is proved that: △ abd ≌ △ Abe, △ ACD ≌ △ ACF. (1 point)  DAB = ≌△ fac, and ∠ BAC = 45 °. [EAF = 90 °. (3 points) and ∵ ad ⊥ BC, ≌≌≌△ Abe, ≌≌△ Abe, ≌≌△ ACF. (1 point) ∵ DAB = EAB, ∠ DAC = fac, and ∠ BAC = 45 °

As shown in the figure, ad ⊥ CD is at point D, BC ⊥ CD is at point C, point E is the midpoint of CD, AE bisection ∠ bad

prove:
As EM ⊥ AB, the vertical foot is m,
∵∠D=∠AME=90°，AE=AE，∠DAE=∠MAE，
In △ ade and △ Ame
∠D＝∠AME
∠DAE＝∠MAE
AE＝AE
∴△ADE≌△AME，
∴DE=EM，
∵DE=EC，
∴EM=EC，
∵EM⊥BE，EC⊥BC，
∴∠MBE=∠CBE，
ν be bisection ∠ ABC

As shown in the figure, ad ⊥ CD is at point D, BC ⊥ CD is at point C, point E is the midpoint of CD, AE bisection ∠ bad

prove:
As EM ⊥ AB, the vertical foot is m,
∵∠D=∠AME=90°，AE=AE，∠DAE=∠MAE，
In △ ade and △ Ame
∠D＝∠AME
∠DAE＝∠MAE
AE＝AE
∴△ADE≌△AME，
∴DE=EM，
∵DE=EC，
∴EM=EC，
∵EM⊥BE，EC⊥BC，
∴∠MBE=∠CBE，
ν be bisection ∠ ABC

It is known that: as shown in the figure, ad ‖ BC, e is the midpoint of line CD, AE bisection ∠ bad Thank you so much. I've already got it. For the first time, some students used the median line and was given a small x.55555 by the teacher

Extend AE and BC and intersect at point F
It is known that ad ‖ BC, ∠ DAE = ∠ BAE, de = EC,
It can be concluded that: ∠ BFA = ∠ DAE = ∠ BAE, AE = EF,
So Ba = BF, be is the center line on the base of the isosceles △ BAF,
It can be obtained that the top angle of be bisection isosceles △ BAF ∠ ABF,
That is: be bisection ∠ ABC