If the areas of △ ABC and △ Dec are equal, and EF = 9, ab = 12, then the value of DF is______ .

If the areas of △ ABC and △ Dec are equal, and EF = 9, ab = 12, then the value of DF is______ .

∵△ ABC and △ Dec have the same area

As shown in the figure, in RT △ ABC, the vertical bisector De of ∠ ACB = 90 ° BC intersects BC at D, AB crosses with E, and F is a point on De （1） Verification: quadrilateral ACEF is a parallelogram; (2) When the size of ∠ B satisfies what condition, the quadrilateral ACEF is diamond? Please answer and prove your conclusion Forget to type a condition AF = CE

(1) ∵ De is the vertical bisector of BC,
Ψ bed = ∠ CED, and de ‖ AC, so e is the midpoint of ab,
In RT △ ABC, CE = AE = be,
Ψ AEF = ∠ AFE, and ∠ bed = ∠ AEF,
∠DEC=∠DFA,
∴AF‖CE,
And ∵ AF = CE,
The quadrilateral ACEF is a parallelogram;
(2) To make the parallelogram ACEF rhombic, AC = CE is enough,
∵CE=1/2 AB,
/ / AC = 1 / 2 ab,
In RT △ ABC, ∠ ACB = 90 °,
When ∠ B = 30 °, ab = 2Ac,
Therefore, when ∠ B = 30 °, the quadrilateral ACEF is rhombic

As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, BC = 3, AC = 4, the vertical bisector De of AB intersects the extension of BC at point E, then the length of CE is () A. 3 Two B. 7 Six C. 25 Six D. 2

∵∠ACB=90°，BC=3，AC=4，
According to Pythagorean theorem, ab = 5,
The vertical bisector De of AB intersects the extension of BC at point E,
∴∠BDE=90°，∠B=∠B，
∴△ACB∽△EDB，
/ / BC: BD = AB: (BC + CE), and BC = 3, AC = 4, ab = 5,
∴3：2.5=5：（3+CE），
Thus, CE = 7
6．
Therefore, B

In the triangle ABC, the bisector ad of the angle cab intersects with the middle perpendicular of BC at point D, DM perpendicular, AB and m, DN perpendicular to AC

It's not over

As shown in the figure, in △ ABC, ∠ C is a right angle. The high Cd on AB and the central line CE exactly divide ∠ ACB. If AB = 20, what are the two acute angles of △ ABC and the values of AD, de and EB?

∵ C is a right angle, CD and CE divide  ACB into three equal parts,
∴∠ACD=∠DCE=∠ECB=1
3×90°=30°，
∵ CD is high,
∴∠A=90°-∠ACD=90°-30°=60°，
∵ CE is the midline,
∴CE=AE=EB=1
2AB=1
2×20=10，
∴∠B=∠ECB=30°，
∴AC=1
2AB=1
2×20=10，
AD=1
2AC=1
2×10=5，
DE=AE=AD=10-5=5．
To sum up: ∠ a = 60 °, B = 30 °, ad = 5, de = 5, EB = 10

As shown in the figure, ∠ ABC = 90 ° in RT △ ABC. Rotate RT △ ABC clockwise for 60 ° around point C to obtain △ Dec, point E on AC, and then flip RT △ ABC along the line of AB by 180 ° to obtain △ ABF. Connect ad (1) The results show that the quadrilateral AFCD is rhomboid; (2) Connect be and extend intersection ad to g, connect CG, what is quadrilateral ABCG special parallelogram and why?

(1) It is proved that RT △ Dec is obtained by the rotation of RT △ ABC around point C by 60 ° and  AC = AF,  ABF = 90 ° and ∵ AC = AF,  ABF = 90 ° and ∵ AC = AF, respectively

As shown in the figure, in RT △ ABC, ∠ ACB = 90 ° and CD is the height above AB edge. If ad = 8 and BD = 2, find CD

∵ in RT △ ABC, ∵ ACB = 90 ° and CD is the height of AB edge
∴∠BDC=∠ACB=90°
∵∠B=∠B
∴△ABC∽△CBD
∴CD2=AD•BD，
∵AD=8，BD=2，
∴CD=
8×2=4．

In the RT triangle ABC, CD is the height on the edge of ab. if ad = 8, BD = 2, find the length of CD To infer the process, it is best to understand the point, the use of Pythagorean theorem, equations... I am too stupid, you science enthusiasts to analyze it

CD^2=AC^2-AD^2
CD^2=AB^2-BC^2-64
CD^2=100-CD^2-BD^2-64
2CD^2=36-4
CD^2=16
CD=4

In the triangle ABC, ab = AC, D, e are the points on BC and AC respectively. When the angle bad and the angle CDE satisfy what conditions, ad = AE, write the reasoning process In the plane rectangular coordinate system, (1) if the coordinates of point a are (1,1), find a point P on the coordinate axis so that △ POA is an isosceles triangle. How many such p points are there? (2) If the coordinates of point a are (2,1), find a point P on the Y axis so that △ POA is an isosceles triangle. How many such p points are there?

A: ad = AE when bad = 2 CDE
It is proved that: ∵ AED = ∵ C + ∠ EDC
∠ADE=∠ADC-∠EDC
=∠B+∠BAD-∠EDC
=∠B+2∠EDC-∠EDC
=∠B+∠EDC
That is ∠ ade = ∠ B + ∠ EDC ②
∵ AB=AC
∴ ∠B=∠C
By comparing the two formulas, we can get the following results
∠ADE=∠AED
∴AD=AE

In the triangle ABC, ab = AC, d = 1 point on BC, e = 1 point on AC, angle bad = 20, and AE = ad, then angle CDE =?

Let ∠ CDE = α, ∠ B = ∠ C = β
∴∠ADE＝∠AED＝∠C＋∠CDE＝α＋β
In addition, ADC = B + bad = β + 20 °
∴∠ADE＋∠CDE＝β＋20°
α＋β＋α＝β＋20°
∴α＝10°