In △ ABC, ab = AC, De is the vertical bisector of AB, the perpendicular foot is D, intersecting AC with E, connecting be. (1) if ∠ Abe = 40 °, calculate the degree of ∠ C; (2) if the perimeter of △ ABC is 82cm, one side is 30cm, calculate the perimeter of △ BCE

In △ ABC, ab = AC, De is the vertical bisector of AB, the perpendicular foot is D, intersecting AC with E, connecting be. (1) if ∠ Abe = 40 °, calculate the degree of ∠ C; (2) if the perimeter of △ ABC is 82cm, one side is 30cm, calculate the perimeter of △ BCE

(1) It is known that ab = AC, De is the vertical bisector of AB, and ∵ a = 40 °

As shown in the figure, in the triangle ABC, ab = AC = 12cm, De is the vertical bisector of AB, intersecting AB, AC at two points D and e respectively. If ∠ C = 70 °, calculate the degree of ∠ CBE and ∠ bec

AB = AC --- > isosceles triangle,
∠C=70°,∠B=70°----->∠A=40°
De is the vertical bisector of AB, so the triangle ade is equal to the triangle BDE
∠DBE=∠A=40°
∠CBE=∠B-∠DBE=30°
∠BEC=180-70-30=80°

As shown in the figure, BD is the angular bisector of ∠ ABC, de ⊥ AB is in E, the area of △ ABC is 30cm2, ab = 18cm, BC = 12cm, then de=______ cm．

As shown in the figure, through point D, make DF ⊥ BC, the perpendicular foot is point F ∵ BD, which is the angular bisector of ∠ ABC, the area of de ⊥ AB, ∵ de = DF ∵ ABC is 30cm2, ab = 18cm, BC = 12cm, ∵ s △ ABC = 12 · de · AB + 12 · DF · BC, that is, 12 × 18 × de + 12 × 12 × de = 30, ∵ de = 2 (CM)

In the triangle ABC, ab = 15, BC = 14, AC = 13, find the area of the triangle ABC

Using Helen's formula: let P = (a + B + C) / 2 = 21
S=√p(p-a)(p-b)(p-c)
=√(21*6*7*8)
=84

In △ ABC, ab = 10cm, BC = 20cm, point P starts from point a and moves along edge AB to point B at a speed of 2cm / s, and point Q starts from point B and moves along edge BC to point C at a speed of 4cm / s. if P and Q start from a and B at the same time, after () seconds, △ PBQ is similar to △ ABC? （　　）
A. 2.5B. 3.5c. 1 and 2.5D. 1 and 3.5

The velocity of point P is 2cm / s, the velocity of point q is 4cm / s, BP = ab-ap = 10-2t, BQ = 4T, ① when BP and ab are corresponding edges, ∵ PBQ ∽ ABC, ∵ bpab = bqbc, that is, 10 − 2t10 = 4t20, the solution is t = 2.5; ② when BP and BC are corresponding edges, ∵ QBP ∽ ABC, ∽ bqab = BPBC, that is, 4t10 = 10 −

In the triangle ABC, angle a = 30 degrees, angle B = 80 degrees, take point D in the triangle to make angle DAC = 10 degrees, angle DCA = 30 degrees, and calculate the degree of angle BDC?
Satisfaction will pursue awards and bonus points

According to the meaning of the title, ∠ ADC = 140, ∠ DAB = 20, ∠ DCB = 40
Let ∠ BDC = x, then ∠ ADB = 360-140-x
From the sine theorem,
In △ BCD, BD / sin40 = BC / SiNx
In △ abd, BD / sin20 = AB / sin (360-140-x)
When BD is eliminated from the simultaneous equations, it can be obtained that,
sin20／sin40＝BCsin（x+140）/〔ABsinx〕
From the sine theorem,
In △ ABC, AB / sinc = BC / Sina
∴sin20／sin40＝sin30sin（x+140）/〔sin70sinx〕
That is sin40sin30sin (x + 140) = sin20sin70sinx
∵sin20sin70＝sin20cos20＝1/2sin40
∴sin（x＋140）＝sinx
∴x＋140＋x＝180
∴x＝20

As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, point D is the midpoint of hypotenuse AB, de ⊥ AC, and the perpendicular foot is e. if de = 2, CD = 25, then the length of be is______ ．

∵ in RT △ ABC, ∵ ACB = 90 °, de ⊥ AC, ∵ de ∥ BC, ∵ point D is the midpoint of AB, de = 2, ∵ BC = 4, ∵ de ⊥ AC, perpendicular foot is e, if de = 2, CD = 25, in RT △ CDE, CE = 4 is obtained by Pythagorean theorem, ∵ in RT △ BCE, ∵ ACB = 90 °, be = BC2 + CE2 = 42

(1) As shown in figure (1), it is known that in △ ABC, ∠ BAC = 90 °, ab = AC, the straight line m passes through point a, BD ⊥ straight line m, CE ⊥ straight line m, and the perpendicular feet are points D and e respectively. Guess the quantitative relationship among de, BD and Ce (just write the result directly). (2) as shown in figure (2), change the condition in (1) to: in △ ABC, ab = AC, D, a and E are all on the straight line m, and have In question (1), is the relationship among De, BD and CE still valid? (3) expansion and application: as shown in figure (3), D and E are two moving points on the straight line m where D, a and E are located (D, a and e do not coincide with each other), point F is a point on the bisector of ∠ BAC, and △ Abf and △ ACF are equilateral triangles, connecting BD and CE. If ∠ BDA = ∠ AEC = ∠ BAC, try to judge the number of DF and ef And explain the reason

(1) De = BD + CE. The reasons are as follows: as shown in Figure 1, ∵ BD ⊥ L, CE ⊥ L, ∵ BDA = ∵ AEC = 90 ° and ∵ BAC = 90 °, ∵ bad + ∵ CAE = 90 °, ∵ bad + ∵ abd = 90 ° and ∵ CAE = ∵ abd & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp

As shown in the figure, in △ ABC, ∠ ACB = 90 °, D and E are two points on the edge of AB, and ad = AC, be = BC. (1) let ∠ a = 60 ° to find the degree of ∠ DCE; (2) let ∠ a = 50 ° to find the degree of ∠ DCE; (3) let ∠ a = a to find the degree of ∠ DCE; (4) please summarize a general conclusion according to the result of solving the problem

Ad = AC, and the \ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\so (1) let ∠ a = 60 ° and find ∠ DCE =(2) set ∠ a = 50 ° to get ∠ DCE = 45 °; (3) set ∠ a = a to get ∠ DCE = 45 °; (4) set ∠ DCE = 45 °

If a is the root of the quadratic equation x ^ 2 + BX + a = O with respect to x, and a is not equal to 0, then the value of B / a can be obtained

Another x1
-B=A+X1
A=AX1
X1=1
-B=A+1
B/A=-（A+1）/A