As shown in the figure, in circle O, AB is the diameter of circle O, CD is a chord, and CD is perpendicular to point P, connecting BC and AD

As shown in the figure, in circle O, AB is the diameter of circle O, CD is a chord, and CD is perpendicular to point P, connecting BC and AD

prove:
Connect AC, BC
Then ∠ ACB = 90 °
∵CP⊥AB
/ / arc BC = arc BD
∴∠A=∠BCP
∵∠CPB=∠CPA =90°
∴△ACP∽△CBP
∴CP/AP= BP.CP
∴CP²=AP*PB

As shown in the figure, in ladder ABCD, ad ∥ BC, ab = ad = DC, AC ⊥ AB, extend CB to f to make BF = CD (1) Find the degree of ∠ ABC; (2) It is proved that △ CAF is isosceles triangle

（1）∵AD∥BC，
∴∠DAC=∠ACB．
∵AD=DC，
∴∠DCA=∠DAC．
∴∠DCA=∠ACB=1
2∠DCB．
∵DC=AB，
∴∠DCB=∠ABC．
∴∠ACB=1
2∠ABC．
In △ ACB, ∵ AC ⊥ ab,
∴∠CAB=90°．
∴∠ACB+∠ABC=90°．
∴1
2∠ABC+∠ABC=90°．
﹤ ABC = 60 °. (3 points)
(2) Proof: connect to DB,
∵ in trapezoidal ABCD, ab = DC,
∴AC=DB．
In the quadrilateral dbfa, Da ∥ BF, Da = DC = BF,
The quadrilateral dbfa is a parallelogram
∴DB=AF，
∴AC=AF．
That is, △ ACF is isosceles triangle. (6 points)

As shown in the figure, fold the rectangle ABCD along AE so that point d just falls on point F on edge BC, ab = 3, ad = 5, and find EF

In the RT triangle ABF, ab = 3, AF = ad = 5
BF^2+AB^2=AF^2
BF=4
CD=AB=3
Let EC = X
EF=DE=CD-CE=3-X
In the RT triangle EFC, EC ^ 2 + FC ^ 2 = EF ^ 2
(3-X)^2=X^2+1^2
X=4/3
EF^2=X^2+1
EF=5/3

As shown in the figure, the rectangle ABCD is folded along AE so that point d falls at point F on the edge of BC. If ad = 5 and ab = 3, then the length of EF is______ .

∵ the rectangle ABCD is folded along AE so that point d falls at point F on the edge of BC,
∴AF=AD=5，EF=DE，
In RT △ ABF, ab = 3,
∴BF=
AF2−AB2=4，
∴CF=BC-BF=5-4=1，
If EF = x, then de = x, EC = 3-x,
In RT △ EFC, ∵ ef2 = EC2 + FC2,
ν x2 = (3-x) 2 + 12, x = 5
3，
That is, the length of EF is 5
3．
So the answer is 5
3．

As shown in the figure, in rectangular ABCD, e is a point on BC, DF ⊥ AE is in F, if AE = BC, find CE = EF

Connecting de
Because ad = BC, AE = BC, so ad = AE,
So angle ade = angle AED
Because ad is parallel to BC
So angle ade = angle CED
So angle AED = angle CED
Because the angle DFE = angle DCE = 90 degrees
So the triangle DFE is equal to the triangle DCE
So CE = EF

In this paper, we give the following results: abed and abef are given Verification: AE bisection ∠ bad

It is proved that: (1) the ∵ 4-quadrilateral, ABCD is a rectangle, ∵ B = ∵ C = ∵ bad = 90 °, ab = CD, ∵ bef + ∵ BFE = 90 ∵ e ⊙ e ⊙ e ⊙ e ⊙ e ∵ e

Radical algorithm Add, subtract, multiply and divide

√a+√b=√b+√a
√a-√b=-(√b-√a)
√a*√b=√(a*b)
√a/√b=√(a/b)

The algorithm that defines the operation '@', is x@y= If the radical is XY + 4x + 4Y, then（ 4@6 ）@8=（ ）

A kind of 4@6=4 ·6+4·4+4·6=64,
∴( 4@6 @ 8=64@8=64 ·8+4·64+4·8=800.

With addition, subtraction, multiplication and division, we can get "24" by calculating the four numbers of 7, 7, 3 and 3. We can only use addition, subtraction, multiplication and division

（3+3/7）*7＝24

Addition, subtraction, multiplication and division algorithm of radical How does (√ 6) + 1 / 1 be converted into 5 (√ 6) - 1 Even if the numerator and denominator are multiplied by the root 6 plus 1, then it's not easy to calculate? By the way, how to add, subtract, multiply and divide the radical

(√ 6) + 1 / 1 numerator and denominator are multiplied by (√ 6) - 1
The denominator becomes 5 and the molecule becomes (√ 6) - 1
Mainly familiar with A2-B2 = (a + b) (a-b)
(a+b)2=a2+2ab+b2
Product, sum and difference, sum difference product
And so on