The triangle ABC, D, e are the points on BC, and BD = EC. It is proved that ab + AC is greater than AD + AE

The triangle ABC, D, e are the points on BC, and BD = EC. It is proved that ab + AC is greater than AD + AE

prove:
Taking BC as the edge, make △ FCB outside △ ABC, so that △ ABC ≌ △ FCB. Connects FD
In △ FDB and △ ace,
DB=EC,
∠DBF=∠ECA,
BF=CA,
So △ FDB ≌ ace
Thus AE = FD
In this way, the inequality becomes Ba + BF > Da + DF in △ AFB
The proof of the latter inequality is as follows
When BF of AD was prolonged to g, AB + BG > Ag = AD + DG;
Because DG + GF > DF,
AB + BG + GF > AD + DF, that is ab + BF > AD + DF

Decomposition factor A ^ 2 + B ^ 2-6a + 6b-2ab + 9

A ^ 2 x B ^ 2-6a x 6b-2ab x 9
=(a²-2ab+b²)-6(a-b)+9
=(a-b)²-6(a-b)+9
=(a-b-3)²

Given that the ellipse C passes through the point a (- 3,2) and has the same focus as the ellipse xsquare + ysquare = 1, the standard equation of ellipse C is obtained?
I'm in a hurry
Is the ellipse x squared / 9 + y squared / 4 = 1

C same
9/（b^2+5）+4/(b^2)=1
b^2=10
Ellipse: x square / 15 + y square / 10 = 1

In triangle ABC, AB > AC, ad is the bisector of angle BAC and intersects BC with D, P is any point of AD, how to prove the following conclusion is correct, that is AB-AC > bp-pc

Take a point E on AB so that AE = AC, connect PE
Easy syndrome △ AEP ≌ △ ACP
So, PE = PC
In △ BPE, there is bp-pe

Let a, X ∈ R, and the complex number x ^ 2 + ax + 1 + 3I is not a pure imaginary number, then the value range of real number a is

Constant is not pure imaginary number
Then x & # 178; + ax + 1 = 0 has no solution
So the discriminant A & # 178; - 4

How to use the first-order function matching method in high school mathematics? F (x + 1) = x2 + X-1, find f (x)
I hope I can write down the process in detail. If I write well, I can add more points

The matching method is to match the right side of the analytic expression with the form in the left bracket
f(x+1)=x²+x-1=[(x+1) -1]²+x+1- 2
Replace x + 1 of the above formula with X, and get
f(x)=(x-1)²+x-2=x²-2x+1+x-2=x²-x -1

The following is a right triangle, with ab as the axis to rotate around it for a circle, get a three-dimensional figure, its volume is how many cubic centimeters?

13 × 3.14 × 62 × 8 = 3.14 × 12 × 8 = 301.44 cubic centimeter; a: its volume is 301.44 cubic centimeter

The resonance condition of RL Series circuit and C parallel circuit is wo = root sign (1 / lc-r ^ 2 / L ^ 2). How did this result come from

When resonant, the voltage and current are in the same phase, that is, the impedance is pure resistance (imaginary part 0). It is more convenient to deduce the parallel circuit with admittance. Similarly, the imaginary part is 0

For a finite set m with n elements, its subsets, proper subsets, nonempty proper subsets and nonempty proper subsets are?

I can't answer what it is, but I can answer how many
There are (n power of 2) subsets, proper subsets [(n power of 2) - 1], nonempty subsets [(n power of 2) - 1], nonempty proper subsets [(n power of 2) - 2]

As shown in the figure, △ ABC is a steel frame, ab = AC, ad is the support connecting a and the midpoint D of BC

In △ abd and △ ACD, ≌ AB = acbd = CDAD = ad, ≌ abd ≌ ACD (SSS)