In △ ABC, ∠ C = 90 °, AC = 4cm, BC = 5cm, point D is on BC, and CD = 3cm. There are two moving points P and Q starting from point a and point B at the same time, where point P moves along AC to terminal C at the speed of 1cm / s, and point Q moves along BC to terminal C at the speed of 1.25cm/s. Passing point P, PE ‖ BC is connected to point E, and EQ is connected (1) The length of AE and De is expressed by the algebraic expression containing x; (2) when the point Q moves on BD (excluding points B and D), let the area of △ EDQ be y (cm2), find the functional relationship between Y and time x, and write out the value range of the independent variable x; (3) when the value of X is, the △ EDQ is a right angle triangle

In △ ABC, ∠ C = 90 °, AC = 4cm, BC = 5cm, point D is on BC, and CD = 3cm. There are two moving points P and Q starting from point a and point B at the same time, where point P moves along AC to terminal C at the speed of 1cm / s, and point Q moves along BC to terminal C at the speed of 1.25cm/s. Passing point P, PE ‖ BC is connected to point E, and EQ is connected (1) The length of AE and De is expressed by the algebraic expression containing x; (2) when the point Q moves on BD (excluding points B and D), let the area of △ EDQ be y (cm2), find the functional relationship between Y and time x, and write out the value range of the independent variable x; (3) when the value of X is, the △ EDQ is a right angle triangle

(1) In RT △ ADC, AC = 4, CD = 3, ∧ ad = 5, ∧ EP ∥ DC, ∧ AEP ∧ ADC, ∧ EAAD = APAC, that is ea5 = x4, ∧ EA = 54x, de = 5-54x When point Q moves on BD for x seconds, DQ = 2-1.25x, then y = 12 × DQ × CP = 12 (4 − x) (2 − 1.25x) = 58 & nbsp; x2 − 72x + 4 (6) the analytic expression of Y and X is y = 58X2 − 72x + 4, in which the value range of independent variable is 0 ＜ x ＜ 1.6. (3) the discussion is divided into two cases: (1) when ∠ eqd = 90 °, it is obvious that EQ = PC = 4-x, and ∵ EQ ∥ AC, ∥ EDQ ∥ ADC ∥ eqac = dqdc, DQ = 1.25x-2, that is, 4 − X4 = 1.25x − 23 The solution is x = 2.5 (9 points) ② when ∠ QED = 90 degree, ∵ CDA = ∠ EDQ, ∵ QED = ∠ C = 90 °∫ EDQ ∫ CDA ∫ dqda = high on the hypotenuse of RT △ EDQ, high on the hypotenuse of RT △ EDQ: 4-x, high on the hypotenuse of RT △ CDA: 125. ∫ 1.25x − 25 = 5 (4 − x) 12, the solution is x = 3.1 (12 points)

In the triangle ABC, the angle c is equal to 90 degrees, CD is perpendicular to AB and D, AC is equal to 6, BD is equal to 8, find DC

CD^2=BD*AD
AC^2=AD^2+CD^2
If AC = 6. BD = 8 is brought in, we can get
CD^2=8AD
36=AD^2+CD^2
The solution is ad = - 4 + 2 radical 13
CD = radical (- 32 + 16, radical 13) = 4 radical (- 2 + radical 13)

As shown in the figure, in ladder ABCD, ad ‖ BC, e is the midpoint of AB, de ⊥ CE

It is proved that the extension line of de-cb is extended to F, ∵ ad ≌ CF, ≌ ad = BF, de = EF, ≁ CE ⊥ DF, ≁ CD = CF = BC + BF, ≌ AD + BC = DC in △ AED and △ bef

In trapezoidal ABCD, ad ∥ BC, ad < BC, e is the midpoint of AB, and de ⊥ CE

Take the midpoint F of DC and connect EF, then EF is trapezoidal median line, then EF = (AD + BC) / 2
And EF is the median line on the hypotenuse of RT △ Dec. EF = CD / 2
So AD + BC = DC

In trapezoidal ABCD, AD / / BC, e is the midpoint of AB, and de ⊥ CE is used to find DC = AD + BC

Extend the CB extension line of De to f
Because AE = be, ead = EBA, EDA = EFB, the triangle AED is all equal to bef
Ad = BF, so as long as FC = CD
Another angle CED = CEF = 90
And EF = EF, EC = EC
So the triangle FEC is all equal to Dec,
So FC = DC, that is AD + BC = DC

Extend the line segment AB to C so that BC = half of AB, D is the midpoint of AC, and DC = 6cm, find the length of AB?
Please write down the solution (what is each step?),

It is known that BC = 1 / 2Ab, ad = 1 / 2Ac, DC = ad = 6cm
So AC = 12cm,
Because AC = AB + BC, BC = ac-ab
So BC = 1 / 3aC = 4cm
So AB = ac-bc = 8cm

As shown in the figure, point C is the midpoint of line AB, point D is the midpoint of line AC, and point E is the midpoint of line BC. 1. If AB = 18cm, find the length of de. 2. If CE = 5cm
And find the length of BD

Analysis: (1) first, AC and BC are calculated from C as the midpoint of line AB, then DC and CE are calculated from D as the midpoint of line AC and E as the midpoint of line BC;
(2) Firstly, the relationship between CE and BD is obtained from (1), and then the length of BD is obtained
(1) ∵ C is the midpoint of ab,
∴AC=BC=1/2AB=9（cm）
∵ D is the midpoint of AC,
∴AD=DC=1/2AC=9/2（cm）
∵ e is the midpoint of BC,
∴CE=BE=1/2BC=9/2（cm）
And ∵ de = DC + CE,
∴DE=9/2+9/2=9（cm）
(2) According to (1): ad = DC = CE = EB,
∴CE=1/3BD
∵CE=5cm,
∴BD=15（cm）

As shown in the figure, point C is the point on line AB, and points D and E are the midpoint of line AC and BC respectively. If AB = 18cm and ad = 4cm, what is CE?
—·———·————·———————·————————·————
A D C E B

CE
=(AB-AC)÷2
=(18-4×2)÷2
=(18-8)÷2
=10÷2
=5cm

Points c, D, e divide line AB into four parts, and AC: CD: de: EB = 2:3:4:5, if CE = 14, find the length of line ab

28

As shown in Figure C, D and E, segment AB is divided into four parts. AC: CD: de: EB = 2:3:4:5 is used. M, P, Q and N are the midpoint of AC, CD, de and EB respectively. If Mn = 21, the length of PQ is calculated
Instead of formulas and equations, it's better to use the relationship between ABCD and mpqn, such as AB + what = what and so on

AC：CD：DE：EB＝2：3：4：5
Let AC = 2B, then CD = 3b, de = 4b, EB = 5b, ab = 14b
If M is the midpoint of AC, then am = AC / 2 = B
If P is the midpoint of CD, then PD = CD / 2 = 3B / 2
If q is the midpoint of De, then DQ = de / 2 = 2B
If n is the midpoint of be, BN = EB / 2 = 5B / 2
MN＝AB－AM－BN＝14b－b－5b/2＝21
Then B = 2 × 21 / 21 = 2
∴PQ＝PD＋DQ＝3b/2＋2b＝7b/2＝7×2/2＝7