In ∠ ABC, AB > AC, ad is high, AE is angle bisector, indicating that ∠ ead = 1 / 2 (∠ C - ∠ b)

In ∠ ABC, AB > AC, ad is high, AE is angle bisector, indicating that ∠ ead = 1 / 2 (∠ C - ∠ b)

In RT ∠ ade, ead = 90 degrees - AED (1)
And ∠ AED is an external angle of ∠ Abe,
therefore
The results show that: AED = ∠ B + ∠ BAE = ∠ B + ∠ BAC / 2 = ∠ B + (180 degrees - ∠ B - ∠ C) / 2 = ∠ B + 90 degrees - ∠ B / 2 - ∠ C / 2 = 90 degrees + ∠ B / 2 - ∠ C / 2,
Then ∠ ead = 90 degrees - (90 degrees + B / 2 - C / 2) = (1 / 2) (∠ C - b)
Given that the positive integer solution of inequality 3x-a ≤ 0 is only 1,2,3, then the value range of a is__
x-a≤0
x≤a/3
The positive integer solutions of ∵ inequality 3x-a ≤ 0 are only 1,2,3
∴x∈[9,12)
If the solution set of inequality 3x-a ≤ 0 is x ≤ A / 3 and the positive integer solution is only 1, 2 and 3, then the value of a must be between 9 and 12. The key is whether the boundary value can be obtained. A = 9 obviously meets the requirements, but a = 12 will produce the fourth positive integer solution, so the value range of a is [9,12]
3x-a≤0,
X
As shown in the figure, AC = AE, ∠ BAF = ∠ bgd = ∠ EAC, is there a triangle congruent with △ Abe in the figure? And prove
In △ ABF and △ DFG, ∠ BAF = ∠ bgd, ∠ BFA = ∠ DFG, ∠ B = ∠ D, ∵ BAF = ∠ EAC, ∵ BAE = ∠ DAC, ∵ AC = AE, ∵ BAE = ∠ DAC, ∠ B = ∠ D, in △ BAE and △ DAC, ∠ B = ∠ D ∠ BAE = ∠ dacae = AC ≌ BAE ≌ DAC (AAS). Answer: Yes. △ BAE ≌ DAC
If the positive integer solution of inequality 3x-a ≤ 0 is 1, 2, 3, then the value range of a is 1______ .
If we solve the inequality 3x-a ≤ 0, we can get x ≤ A3. The positive integer solution of ∵ inequality is 1, 2, 3, 3 ≤ A3 < 4, and the solution is 9 ≤ a < 12. So the answer is: 9 ≤ a < 12
As shown in the figure, in △ Abe, ab = AE, ad = AC, ∠ bad = ∠ EAC, BC and de intersect at point o
In △ ABC and △ AED, in △ ABC and △ AED, ab = AE in △ ABC and △ AED, ab = AE, in △ ABC and △ AED, ab = AE, in △ ABC and △ AED, ab = AE. In △ ABC and △ AED, ab = AE, ab = AE, in △ ABC = eadac = ad, and △ ABC △ AED (eaddac) (as) in @ (2) by (2) knowing (1) from (1) from (1) from (1) knowing (1) knowing 2
Given that the solution of the equation x − 42-a + 1 = x is suitable for the inequalities - 12x ≤ - 1 and X-2 ≤ 0, the value of a is obtained
The solution set of inequality-12x ≤ - 1 is x ≥ 2, the solution set of inequality X-2 ≤ 0 is x ≤ 2, so the value of X satisfying both inequalities is x = 2. Substituting x = 2 into x − 42-a + 1 = x, a = - 2 is obtained
As shown in the figure, AC = AE, ∠ BAF = ∠ bgd = ∠ EAC, is there a triangle congruent with △ Abe in the figure? And prove
In △ ABF and △ DFG, ∠ BAF = ∠ bgd, ∠ BFA = ∠ DFG, ∠ B = ∠ D, ∵ BAF = ∠ EAC, ∵ BAE = ∠ DAC, ∵ AC = AE, ∵ BAE = ∠ DAC, ∠ B = ∠ D, in △ BAE and △ DAC, ∠ B = ∠ D ∠ BAE = ∠ dacae = AC ≌ BAE ≌ DAC (AAS). Answer: Yes. △ BAE ≌ DAC
It is known that y is the inverse proportional function of X, and when x = 4, y = 6, if the value range of the independent variable x is 2 greater than or equal to x less than or equal to 3
The range of choosing a position for y
y=k/x
6=k/4
k=24
y=24/x
2=
As shown in the figure, in △ ABC, ∠ BAC = 120 ° ad ⊥ BC is in D, and ab + BD = DC, then ∠ C=______ Degree
Cut off de = dB on DC, connect AE, let ∠ C = x, ∵ AB + BD = DC, de = dB, ∵ CE = AB, and ∵ ad ⊥ BC, DB = De, ∵ straight line ad is the vertical bisector of be, ∵ AB = AE, ∵ CE = AE, ∵ B = ∠ AEB, ∵ C = ∠ CAE, and