As shown in the figure, in △ ABC, ad bisects ∠ BAC, de ‖ AC, EF ⊥ AD and BC extension line to F

As shown in the figure, in △ ABC, ad bisects ∠ BAC, de ‖ AC, EF ⊥ AD and BC extension line to F

It is proved that: ∵ ad bisection ∵ BAC, ∵ bad = ∵ CAD, ∵ de ∥ AC, ∵ EDA = ∵ CAD, ∵ EDA = ∵ ead, ∵ AE = ed, and ∵ EF ⊥ ad, ∵ EF is the vertical bisection line of AD, ∵ AF = DF, ∵ fad = ∵ FDA, and ∵ fad = ? CAD + ∵ fac, ∵ FDA = ∵ B + ∵ bad, ∵ fac = ? B
If the solution of the equation AX + 12 = 0 about X is 3, then the solution of inequality (a + 2) x > - 8 is x?
Finding the solution of X
Substituting x = 4 into ax + 12 = 0
ax=-12
a=-3
So (a + 2) x > - 8
-x>-8
X
Substituting x = 3 into ax + 12 = 0
ax=-12
a=-4
So (a + 2) x > - 8
-2x>-8
X-8, so X-8
-2x>-8
X-8
(﹣4+2)x>﹣8
﹣2x>﹣8
X
As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, AC = BC, D is the midpoint on the edge of BC, CE ⊥ ad at point E, BF ∥ AC intersects the extension line of CE at point F
It is proved that in ∵ RT △ ABC, ∵ ACB = 90 °, AC = BC, ∵ BF ∥ AC, ∵ ACB = CBF = 90 °, ∵ CE ⊥ ad, ∵ 2 + 3 = 90 °, ∵ 1 = 3, in ? ACD and ≌ CBF, ? ACD ≌ CBF, ≌ BF = CD, ∫ D is the midpoint on the edge of BC, ∫ BD = CD, ≌ BD = BF
The solution of inequality 3 (x-1) > 2 (x 1) is the solution of inequality ax > B. what is the relationship between a and B
The first inequality is a x > 2
Solve the second inequality as B x > b / a (a > 0) (1)
X
As shown in the figure, ad is the bisector of ∠ BAC, de ⊥ AB, e for perpendicular foot, DF ⊥ AC, f for perpendicular foot, and BD = CD
It is proved that: ∵ ad is the bisector of ∠ BAC, de ⊥ AB, DF ⊥ AC, ∵ de = DF, and ∵ BD = CD, ≌ RT △ DBE ≌ RT △ DCF (HL). ∵ be = CF
If the solutions of inequality 3 (x-1) > 2 (x + 1) are all solutions of inequality ax > b, what relationship should a and B satisfy?
Solve inequality 3 (x-1) > 2 (x + 1), remove bracket, get: 3x-3 > 2x + 2, transfer term, get: 3x-2x > 3 + 2, merge similar term, get: x > 5. The coefficient of inequality ax > b is changed into 1, then both sides divide by a at the same time, then the relationship between a and B is: BA ≥ 5, and a > 0
As shown in the figure, ad is the bisector of ∠ BAC, de ⊥ AB, e for perpendicular foot, DF ⊥ AC, f for perpendicular foot, and BD = CD
It is proved that: ∵ ad is the bisector of ∠ BAC, de ⊥ AB, DF ⊥ AC, ∵ de = DF, and ∵ BD = CD, ≌ RT △ DBE ≌ RT △ DCF (HL). ∵ be = CF
If a ＞ - B ＞ 0, the solution set of the inequality system ax ＞ BBX ＞ a about X is ()
A. BA ＜ x ＜ abb. No solution C. x ＞ bad. X ＞ ab
If ∵ - B ∵ 0 and ∵ B ∵ 0, the solution of inequality (1) is x ∵ Ba, and the solution of inequality (2) is x ∵ AB and ∵ BA ∵ ab. the system of inequalities has no solution
As shown in the figure, in △ ABC, ∠ C = 90 °, ad is the bisector of ∠ BAC, de ⊥ AB is on e, f is on AC, BD = DF. The following results are obtained: (1) CF = EB; (2) CBA + ∠ AFD = 180 °
(1) In RT △ DCF and RT △ DEB, BD = dfdc = De, at △ DCF ≌ RT ≌ DEB (HL), CF = EB. (2) ≔ RT ≌ DCF ≌ RT ≌ DEB, (2) ≂ RT ≌ DFC = ∠ B. ≂ DFC + ∠ AFD = 180 ° and ≌ cab + ∠ AFD = 180 °
If a > - b > 0, then the system of inequalities ax > b, BX about X
If a > - b > 0,
On the system of inequalities of X:
ax>b,x>b/a
bxb>a
b/a>a/b
On the inequality system ax > b, BXB / a of X
The solution of ax > b is x > b / A,
BXA / b