As shown in the figure, in △ ABC, D is the midpoint of BC, de ⊥ AB is at e, DF ⊥ AC is at F, and be = CF

As shown in the figure, in △ ABC, D is the midpoint of BC, de ⊥ AB is at e, DF ⊥ AC is at F, and be = CF

Known de ⊥ AB, DF ⊥ AC
Therefore, ∠ bed = ∠ CFD = 90 °
It is known that D is the midpoint of BC, so BD = CD
Known be = CF
So, RT △ bed ≌ RT △ CFD (HL)
Then, B = C
That is, △ ABC is an isosceles triangle
And D is the middle point of bottom BC
So, ad ⊥ BC
If de ⊥ AB, DF ⊥ AC, then angle bed = angle CFD = 90 degrees
D is the midpoint of BC, then BD = CD
Because be = CF
So triangle BDE is congruent with triangle CDF, so angle B = angle C
So AB = AC, △ ABC is an isosceles triangle, and D is the midpoint of BC
So ad, BC
∵ D is the midpoint of BC
∴BD=DC
⊙ de ⊥ AB at e, DF ⊥ AC at point F
∴∠BED=∠CFD=90°
∵BE=CF
∴△BED≌△CFD
∴DE=DF
∵DA=DA
∴△AED≌△AFD
∵△BED≌△CFD
∴△ABD≌△ACD
∴AD⊥BC
Because point D is the midpoint of BC, BD = DC
Because DEB = 90 degrees, de ⊥
DF ⊥ AC, so the angle DFC = 90 degrees,
Because be = CF
So △ BDE is all equal to △ DFC, so angle B = angle C
It is concluded that △ ABC is an isosceles triangle with ab = AC
And because point D is the midpoint of BC,
So ad ⊥ BC
∵BE=CF,∠BED=∠CFD=90º,BD=CD，
≌ triangle bed ≌ triangle CFD,
∴BE＝CF，DE=DF；
∵ ad = ad, ∠ DEA = ∠ DFA = 90 & # 186;,
≌ triangle DEA ≌ triangle DFA,
∴AE=AF
∴AB＝AC
The triangle ABC is isosceles triangle,
∴AD⊥BC。
Discuss the function f (x) = x ^ 2,0
When 0
In the sequence {an}, if A1 = A2 = 1 and an + 2 = an + 1 + an (n ∈ n ^ 2), it is proved by mathematical induction that a 5N can be divided by 5
When n = 1, A3 = a1 + A2 = 2a4 = A3 + A2 = 3A5 = A3 + A4 = 5 satisfies the assumption that n = k satisfies a (5K + 5) = a (5K + 4) + a (5K + 3) = 2A (5K + 3) + a (5K + 2) = 2 [a (5K + 2) + a (5K + 1)] + a (5K + 1) + a (5K) = 2 [a (5K + 1) + a (5K) + a (5K + 1)] + a (5K + 1) + a (5K a (5K) = 5A (5K + 1) + 3A (5K 5a) 5A (5K + 1) can be replaced by 5
Let a (5N) be divisible by 5
Then a (5N + 5) = a (5N + 3) + a (5N + 4) = a (5N + 1) + 2A (5N + 2) + a (5N + 3)
=a(5n+1)+2a(5n)+2a(5n+1)+a(5n+1)+a(5n+2)
=2a(5n)+4a(5n+1)+a(5n)+a(5n+1)
=3a(5n)+5a(5n+1)
First, 3A (5N) must be divisible by 5
In addition, the sequence is expanded by
Let a (5N) be divisible by 5
Then a (5N + 5) = a (5N + 3) + a (5N + 4) = a (5N + 1) + 2A (5N + 2) + a (5N + 3)
=a(5n+1)+2a(5n)+2a(5n+1)+a(5n+1)+a(5n+2)
=2a(5n)+4a(5n+1)+a(5n)+a(5n+1)
=3a(5n)+5a(5n+1)
First, 3A (5N) must be divisible by 5
In addition, the sequence is all integers, so 5A (5N + 1) can also be divided by 5
It is proved that a (5N + 5) can be divisible by 5
Obviously A5 = 5 can be divided by 5, so A10, A15... Can be divided by 5
Question: just tell me the general formula of an
When we ask the value of a, the linear equations ax1 + x2 + X3 = 1, X1 + AX2 + X3 = a, X1 + x2 + AX3 = a ^ 2 have unique solutions, no solutions and infinite solutions?
Refer to this: when λ takes what value, the system of non-homogeneous linear equations has unique solution, no solution, infinite solution, λ X1 + x2 + X3 = 1x1 + λ x2 + X3 = λ X1 + x2 + λ X3 = λ ^ 2. The augmented matrix is λ 1111 λ 1111 λ ^ 2. First calculate the determinant of coefficient matrix, λ 1111 λ = (λ + 2) (λ - 1) ^ 2
As shown in the figure, △ ABC, be and CF are the heights on the sides of AC and ab respectively, and D is the midpoint on the side of BC
emergency
The hypotenuse is half the center line of a right triangle
De FD is the midline of RT △ BCD and RT △ BFC
So de = BC / 2 FD = BC / 2 so FD = de
What is the type of function discontinuity of y = (1 + x) / (2-x ^ 2),
Y = (1 + x) / (2-x ^ 2), the function discontinuity point x = ± √ 2, the left and right limit of function at them do not exist, so it is the second kind of discontinuity point
It is known that the sequence {an} satisfies an + 1 = an − 22An − 3, n ∈ n *, A1 = 12. (I) calculate A2, A3, A4; (II) guess the general term an of the sequence, and prove it by mathematical induction
(I) from the recurrence formula, A2 = A1 − 22a1 − 3 = 12 − 22 · 12 − 3 = 34, (3 points) (II) conjecture: an = 2n − 12n. (5 points) prove: ① when n = 1, the equation holds. (6 points) ② when n = K (K ∈ n *), the equation holds. That is, AK = 2K − 12K. (7 points) AK + 1 = AK
When a is the value of a, the system of linear equations: ax1 + x2 + X3 = a + 2, X1 + AX2 + 2x3 = 4, 2x1 + 2x2 + AX3 = a ^ 2 + 4 has unique solutions and infinite solutions,
The coefficient determinant | a | = a 111a 222ac3-c2a 101A 2-A2 2a-2r2 + R3a 103a + 202 2a-2 = (A-2) [a (a + 2) - 3] = (A-2) (a ^ 2 + 2a-3) = (A-2) (A-1) (a + 3). So when a ≠ 1 and a ≠ 2 and a ≠ - 3, the equations have unique solutions
It is known that: as shown in the figure, D is the midpoint of BC side of △ ABC, de ⊥ AB, DF ⊥ AC, the perpendicular feet are e, f respectively, and de = DF. Please judge what triangle ⊥ ABC is? And explain the reason
It is proved that connecting ad, ∨ de ⊥ AB, DF ⊥ AC, ∨ bed = ∠ CFD = 90 ° and de = DF, ∨ D is the midpoint on the BC side of △ ABC, ∨ BD = DC, ≌ RT △ EBD ≌ RT △ FCD (HL), ∨ EBD = ∠ FCD, ≌ ABC is an isosceles triangle
The continuity of function is discussed. If there is a discontinuity, it is shown that the type of discontinuity (Sint / SiNx) ^ (1 / (t-x)) t approaches X
Where's a function here? It's a limit
Forget to add "f (x) =" before?
This is an indeterminate formula of 1 ^ infinite power type. First, use the second important limit to find the expression of the function
f(x)=e ^ [cosx /sinx]
The discontinuity is x = k π (k = 0, plus or minus 1, plus or minus 2.)
infinite discontinuity