Determinant calculation, into a triangle Find out how the following determinant is transformed into a triangle 0 a b a a 0 a b b a 0 a a b a 0

r3-r1,r4-r2
0 a b a
a 0 a b
b 0 -b 0
0 b 0 -b
In line 3 and 4, B is proposed respectively
0 a b a
a 0 a b
1 0 -1 0
0 1 0 -1
r2-ar3,r1-ar4
0 0 b 2a
0 0 2a b
1 0 -1 0
0 1 0 -1
r1r3,r2r4
1 0 -1 0
0 1 0 -1
0 0 b 2a
0 0 2a b
Determinant = B ^ 2 (b ^ 2-4a ^ 2) = B ^ 4-4a ^ 2B ^ 2

How to calculate the algebraic covalent of determinant elements?

In n-order determinant det (a), after the i-th row and j-th column of the element AIJ ((I, J) is the subscript, the same below), the N-1 order determinant composed of the remaining elements in the original order is called the cofactor of element AIJ, which is counted as mij, and the (I + J) power of AIJ = - 1 is multiplied by mij to be the algebraic cofactor of element AIJ

Calculate the algebraic cofactor of the elements in the third row of determinant, and find out each determinant
The first line: 1, - 1,0,1
The second line: 2,0,2, - 1
The third line: A, B, C, D
The fourth line: 3,1,3, - 2

A31= -1
A32 = 1
A33 = 2
A34 = 2
D = b - a + 2c + 2d

In △ ABC, D is the point on AB, AE ⊥ CD, the perpendicular foot is e, ad = 2, DB = 1, AC = root 6, and ∠ ACB = 60 °, find the length of AE and the degree of ∠ ACD

∵ AD / AC = AC / AB = 2 / √ 6 (= √ 6 / 3), and ∠ DAC = ∠ cab
∴ △DAC∽△CAB ∴ ∠CDA=∠BCA=60°
The AE = ad × sin60 ° = 2 ×√ 3 / 2 = √ 3
In the case of case, sin ∠ ACD = AE / AC = √ 3 / √ 6 = √ 2 / 2
∴ ∠ACD=45°

In △ ABC, D is a point on the edge of AB, AE ⊥ CD, ad = 2, DB = 1, AC = root 6, ∠ ACB = 60 °. Find the length of AE and the degree of ∠ ace?

As shown in the figure, in △ ABC, ad is the middle line on the edge of BC, f is a point on the edge of AD, and aeeb = 16, and the ray CF intersects AB at point E, then AFFD is equal to___ ．

As shown in the figure: through point D, make DG ‖ EC intersect AB with G, ∵ ad is the midline on the edge of BC, ∵ GD is the median line of △ BEC, ∵ BD = CD, BG = Ge. ∵ aeeb = 16, ∵ AEEG = 13 ∵ DG ‖ EC, ∵ AEEG = AFFD = 13

Are there triangles congruent with △ Abe in the graph? Please prove your conclusion

Angle B = D ∠ BAM = EAC so that angle BAE = angle CAD and AC = AB triangle ACD are all equal to △ Abe

As shown in the figure △ Abe, ab = AE, ad = AC, angle bad = angle EAC, BC, ed intersect with point o

(1) It is proved that because ∠ bad = ∠ EAC, then ∠ BAC = ∠ ead, and because AB = AE, AC = ad, so △ ABC ≌ △ AED. (2) because △ ABC ≌ △ AED, so ∠ ABC = ∠ AED, and because AB = AE, so ∠ Abe = ∠ AEB, so ∠ Abe - ∠ ABC = ∠ AEB - ∠ AEB, that is ∠ CBE = ∠ DEB

In triangle ABC, the bisector ad of angle a intersects B C with D, and it is proved that BD ratio DC equals AB ratio a
Great Xia, please
C

Make de ‖ AC cross AB to e through point D
∴BD:DC=BE:AE
∠EDA=∠DAC=∠DAE
△BDE∽△BCA
∴AE=DE,BE:DE=AB:AC
∴BD:DC=BE:DE=AB:AC
That is BD: CD = AB: AC

As shown in the figure, in △ ABC, BD ⊥ AC, CE ⊥ AB, the perpendicular feet are D, e, BD, CE, respectively, intersecting at the point o [1] ∠ a = 50 ° to find the relationship between ⊥ BOC [2] ∠ BOC and ⊥ a

(1) The reasons are as follows: ∵ BD ⊥ AC, CE ⊥ ab ∪ CDO = ∠ AEC = 90 °∠ DCO = 90 ° - ∠ a ∫ BOC = ∠ CDO + ∠ DCO ∪ BOC = 90 ° - ∠ a + 90 ° = 180 ° - ∠ a ∪ a = 50 °∠ BOC = 130 ° (2) ∠ BOC = 180 ° - ∠ a

Determinant calculation, into a triangle Find out how the following determinant is transformed into a triangle 0 a b a a 0 a b b a 0 a a b a 0