Three geometric proofs In the square ABCD, D is de ‖ AC, the angle ace is 30 ° and CE is crossed with ad to F. it is proved that AE = AF

Three geometric proofs In the square ABCD, D is de ‖ AC, the angle ace is 30 ° and CE is crossed with ad to F. it is proved that AE = AF

It is proved that if BD is connected, AC is crossed to o, and E is used as △ ace, the edge AC is higher eg,
In the square ABCD, do = Bo = BD / 2 = AC / 2,
Because de ‖ AC
So eg = do
Because in the right angle △ CEG, eg = CE / 2,
So AC = EC,
Because ∠ ace = 30 °,
So ∠ e = (180-30) / 2 = 75,
Because ∠ EFA = ∠ DAC + ∠ ace = 75
So ∠ AEF = ∠ EFA
So AE = AF

Geometric proof in junior high school It is known that △ ABC has an angle with P as its vertex, and ∠ P = half ∠ ACD. If the vertex of this angle is placed on BC, one side of the angle always passes through point a, and the other side intersects with the bisector of the outer angle of ∠ ACD at point E (2) As shown in Fig. 2, when ∠ ACB = 90 ° and Ca = CB, what is the quantitative relationship between segments CP, CE and AC?

I just answered your question in another post
After simplification, AP = (radical 2) * PE
And I found a better way,
Angle ACP = 90 degrees, so C is on a circle with AP as its diameter
Since angle PAC = angle PEC, e is also on the circle determined by APC
Therefore, the angle AEP = 90 degrees is the conclusion

Square proof questions, In the square ABCD, P and Q are the points on BC and CD respectively. If the circumference of triangle PCQ is equal to half of the circumference of square, try to explain the angle PAQ = 45 degrees!

Suppose the side length of a square is 1, BP = a, DQ = B, then PQ = a + B, 0 〈 = a 〈 = 1,0 〈 = B 〈 = 1 because the circumference of the triangle PCQ is equal to half of the circumference of the square, PQ = BP plus DQ. Because PQC is a right triangle, the square of PC + the square of QC = the square of PQ, that is, the square of (1-A) + the square of (1-B) = (a + b)

P is any point of the edge CD of the square ABCD, BG is perpendicular to AP, take point E on AP, make Ag = Ge, connect be and CE 1) Confirmation: be = BC 2） If the bisector of angle CBE intersects the extension line of AP at point N, connect DN, and verify: BN+DN= AN with root sign of 2 times (the most important question is the second question ~)

It is proved that: (1) square ABCD, ab = BC = CD = da
∵ BG⊥AE,AG=GE,Rt△ABG≌Rt△BGE
∴ AB=BE=BC
Connect CN, extend BN and CE to H
It is obvious that RT △ ADM ≌ rtabg, DM = AG
∵ BN bisection  CBE, ᙽ ch = he
∵ ∠CBN=∠EBN,BE=BC,BN=BN
Ψ BCN ≌ △ Ben, ᚔ CN = ne, △ CEN is isosceles △
If the extension of AE AC DC extension line is at F, then there is: ∠ bag = ∠ Berg = ∠ CFE = ∠ BCN
A. B, C, D, n five points are in the same circle, ∠ and = ∠ bng = 45 ° [the circumference angle of the circle to which the AB chord is aligned = 45 °]
RT △ DMN and RT △ BGN are isosceles right triangle, √ 2DM = √ 2ag = DN, √ 2gn = BN, √ 2ag + √ 2gn = √ 2An = BN + DN
The standard answer is not to do any auxiliary line, only isosceles triangle and right triangle through
It is concluded that RT △ BPG is an isosceles right triangle
Then, am = GN
reference resources:
⑴ ⊿BGA≌⊿BGE（SAS）,BE=BA=BC
⑵ ⊿BNC≌⊿BNE（SAS）,∴∠BCN＝∠BEN＝∠BAE.
A. B, C, D, n are in the same circle. DNB = 90 °. Make the vertical line AK of an intersect the extension line of Nd at K
≌ ADK ≌ ABN, DK ≌ ADK ≌ ⊿ ABN, DK = BN.AN=AK
⊿ ank is an isosceles right triangle, BN + DN = KD + DN = kn = √ 2An

Proof of plane geometry There is an isosceles trapezoid whose top edge is equal to the waist. It is proved that the top edge is half of the bottom edge

The title is wrong. It seems that there is no condition
The result is valid only when the base angle is 60 degrees

Junior high school geometry proof square proof problem is easy, but there is no way, urgent? The distance between the point E on the edge of the square ABCD, BC and the point F on the outer diagonal of the angle c = EA, and the verified angle FEA = 90 degrees,

But the principle of this problem is that in an obtuse triangle, edges and corners are OK
If be = BG, Ag = EC (both be and BG are subtracted by the side length of the square) AE = EF ∠ EGA = ∠ ECF = 135 °
And then we can get the congruence

Given that the diameter of the circle is 13cm and the distance from the center of the circle to the straight line L is 6cm, then the number of common points between the line L and the circle is______ .

The radius of the circle is 6.5cm
Because the distance between the center of the circle and the line L is 6cm, the line and the circle intersect, so there are two intersections

On the geometry of the circle in the third day of junior high school! Given that the distance between point O and line L is 3, on the circle with point o as the center, only the distance between two points and line L is 1. What is the range of radius length r of this circle?

When the radius is less than 2, the distance from all points to L is greater than 1, which is very easy to understand. 2. When the half diameter is equal to 2, only the nearest point to L is the distance

A square is inscribed in a circle with a radius of 1 cm Hurry!

Diameter of circle = diagonal length of square
So diagonal = 1 × 2 = 2 cm
Let the side length of a square be a
Using the Pythagorean theorem, 2A 2 = 4,
a²=2.
Square area = 2 square centimeter

The area of a square is 23, and there are four circles in the square, whose radius is r ① Find the radius of each circle. ② if the square area is expanded by 4 times, how many times does the radius of each circle increase

To provide you with a detailed problem-solving steps, I hope to help you!
(1) The side length of the square is √ 23
Then r = √ 23 / 4
The area of the circle is:
S=π*r²=3.14*23/16=4.51
(2) If the square area is increased by four times, the side length will be enlarged
√ 4 = 2 (Times)
The radius of the circle is
r=√（4*23）/4=√23/2
And √ 23 / 2 △ √ 23 / 4 = 2
So the radius is doubled
Glad to have the opportunity to answer for you, if you do not understand, welcome to ask, happy every day!