Calculate the value of the determinant 31-12-513-4201-11-53-3, This is called four section determinant

Can we use determinant to expand the theorem by row and column? & nbsp; & nbsp; if you haven't learned it yet, just swap columns 1 and 2, and the calculation after that will be clear

Change this determinant into triangle determinant and evaluate it
1 1 2 3
1 2 3 -1
1 -1 -1 -2
2 3 - 1 - 1 please write the process

r2-r1,r3-r1,r4-2r1
1 1 2 3
0 1 1 -4
0 -2 -3 -5
0 1 -5 -7
r3+2r2,r3-r2
1 1 2 3
0 1 1 -4
0 0 -1 -13
0 0 -6 -3
r4-6r3
1 1 2 3
0 1 1 -4
0 0 -1 -13
0 0 0 75
Determinant = - 75

The determinant is transformed into upper triangular form
0 1…… 1 1
1 0…… 1 1
……………………
1 1…… 0 1
1 1…… 1…… 0

Add each line to line 1,
obtain
n-1 n-1…… n-1 n-1
1 0 …… 1 1
……
1 1 …… 1 0 extracts n-1 from line 1
=
1 1…… 1 1 *(n-1)
1 0…… 1 1
……
1 1…… 1 0 subtract line 1 from each line
=
1 1…… 1 1 *(n-1)
0 -1…… 0 0
……
0 0…… 0 - 1 so we get the triangle determinant
=
(n-1)* (-1)^(n-1)

It is known that be CF is the angular bisector of triangle ABC. Be CF intersects D. if angle a = 50 degrees, then angle BDC
It is known that be and CF are bisectors of triangle ABC. Be and CF intersect at D. if angle a = 50 degrees, then angle BDC=

It is known that be and CF are bisectors of triangle ABC, so angle BCD + angle CBE = (180 ° - 50 °) △ 2 = 65 °
Angle CBE = 180 ° - 65 = 115 °

As shown in the figure, in the equilateral triangle ABC, the bisectors of ∠ ABC and ∠ ACB intersect at points o, and the vertical bisectors of Bo and OC intersect at points E and f respectively
Prove that △ OEF is an equilateral triangle

Because the bisector of ∠ ABC and ∠ ACB intersects at point O, ∠ OBC = 30 ° and ∠ OCB = 30 °
Because the vertical bisectors of Bo and OC intersect BC at points E and f respectively, ∠ EOB = ∠ OBC = 30 ° and ∠ FOC ∠ OCB = 30 °
Therefore, OEF = 60 °∠ ofe = 60 °,
Delta OEF is an equilateral triangle

In the equilateral triangle ABC, the bisectors of ∠ ABC and ∠ ACB intersect at point O, and the vertical bisectors of Bo and OC intersect BC at point E and f respectively, indicating that △ OEF is an equilateral triangle
Equilateral triangle

∵EB=EO =>∠EBO=∠EOB=1/2∠ABC=30° =>∠OEF=∠EBO+∠EOB=60°
In the same way, if ∠ ofe = 60 °, the ∧ OEF is an equilateral triangle

Given the equilateral △ ABC, as shown in the figure, the bisectors of angle B and angle c intersect o, and the vertical bisectors of Bo and co intersect BC at point EF respectively. Can you get BF = EF = FC?
Please give reasons

Yes
Firstly, we connect EO and fo, because they are vertical bisectors, we can get BeO and CFO are isosceles triangle
So be = EO CF = fo
Because ABC is an equilateral triangle, Bo and Co are bisectors
Therefore, OBE = BOE = FOC = OCF = 30 degree
Therefore, OEF = ofe = 60 degree
So the triangle OEF is an equilateral triangle
So OE = of = EF
Because be = EO, CF = fo (before)
So be = EF = FC

In equilateral △ ABC, the bisectors of ∠ ABC, ∠ ACB intersect at point O. if the vertical bisectors of OB and OC intersect at points E and F, the quantitative relationship between EF and ab is conjectured and proved

It is proved that: if the bisector connecting OE, of. ∵ OB and OC intersects BC at point E, f ∵ OE = EB, of = FC ∵ △ ABC is an equilateral triangle, and the bisector of angle ABC and angle ACB intersects o ∵ OBE = ∵ OEB = 30 ° and ∵ OEF = 60 ° and ∵ ofe = 60 ∵ OEF is an equilateral triangle ∵ be = EF = FC ∵ EF = 13ab,

Ao is the angular bisector of triangle ABC (AB > AC), and the vertical bisector of ad intersects the extension of BC at E. let CE = a, de = B, be = C
Ao is the angular bisector of the triangle ABC (AB > AC), the vertical bisector of ad intersects the extension of BC at e, let CE = a, de = B, be = C, prove that there are two real roots for the quadratic equation of one variable a of X multiplied by the square of X minus 2bx plus C equal to 0
Pattern: there is a big triangle outside, which is Abe (from left to right). CE looks like AB, de equals AE

It is proved that: connecting AC, because the vertical bisector of ad intersects the BC extension line at e, so de = AE, so ∠ ade = ∠ DAE, in △ abd, ∠ ade = ∠ B + ∠ bad, ∠ DAE = ∠ DAC + ∠ CAE, because ad bisectors ∠ BAC, so ∠ bad = ∠ CAD, so ∠ B = ∠ CAE, and ∠ ACB is the common angle, so CE / AE = AE / b of △ ace ∽ BAE

In the equilateral triangle ABC, ad is the center line on the edge BC. In the equilateral triangle ade, De is called AC and point F. is AC the vertical bisector of de? Why

Because the triangle ABC is an equilateral triangle
So AB = AC, BAC = 60 degrees
And because ad is the middle line of BC
So the angle bad = angle DAC = 30 degrees
So AC bisector DAE
Because the triangle ade is an equilateral triangle
So the angle ade = 60 degrees
So EDC = 30 degrees
So the angle DFC = 90 degrees
So AC splits de vertically

Calculate the value of the determinant 31-12-513-4201-11-53-3, This is called four section determinant