As shown in the figure, both the triangle ABC and the triangle Dec are isosceles right triangles. The angle ACB = angle DCE = 90 ° f is the midpoint of De, h is the midpoint of AE, and G is the midpoint of BD If the △ Dec in Fig. 1 is rotated clockwise by an acute angle around point C to Fig. 2, is the triangle FGH still an isosceles right triangle

As shown in the figure, both the triangle ABC and the triangle Dec are isosceles right triangles. The angle ACB = angle DCE = 90 ° f is the midpoint of De, h is the midpoint of AE, and G is the midpoint of BD If the △ Dec in Fig. 1 is rotated clockwise by an acute angle around point C to Fig. 2, is the triangle FGH still an isosceles right triangle

I will not elaborate (let FG and AD be handed over to I)
Because F, h and G are the midpoint of ED AE BD, FG and HF are the median lines of the triangle DEB ead, so FG ‖ EB HF ‖ ad FG is half of be and HF is half of AD. according to two isosceles right triangles, ad = EB is obtained, so HF = GF angle HFG = AIG = ACB = 90, so triangle HFG is isosceles straight angle triangle
Secondly, we can prove that be ad is similar to the method just now

It is known that △ ABC and △ Dec are equilateral triangles, ∠ ACB = ∠ DCE = 60 ° and B, C and E are on the same line, connecting BD and AE. It is proved that DF = Ge

It is proved that ∵ ABC and △ Dec are equilateral triangles  BC = AC, CD = CE, ∠ BCA = ∠ DCE = 60 °∠ BCD = ∠ ace = 60 ° + ∠ ACD ? BCD congruence △ ace ?

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 hypotenuse ab. if ad = 8 and BD = 2, find CD Can you make it clear?

cd=4
There is such a formula
The square of the height of the hypotenuse = the product of the two parts of the hypotenuse
Do you understand? In fact, the principle of this formula is similar to a triangle

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．

As shown in the figure, in the triangle ABC, ab = AC, points D and E are on the extension line of BC and AC respectively, ad = AE, angle CDE = 30 degrees, (1) if the angle B = x, use the algebraic formula containing x to express the angle E. (2) calculate the degree of angle bad

1. Because AB = AC, angle B = angle ACB = x because D, e is on the extension line of BC and AC, so angle ACB = angle DCE = x, so angle e = 180 ° - X-30 ° = 150 ° - X2, because ad = AE, angle ade = angle e = 150 ° - x angle ead = 180 ° - 2 (150 ° - x) because AB = AC, angle BAC = 180 ° - 2x, so angle bad = angle BAC + angle ead =

As shown in the figure, it is known that △ ABC and △ CDE are equilateral triangles. It is proved that BD = AE

prove:
∵ △ ABC and △ CDE are equilateral triangles
∴BC=AC,CD=CE,∠ABC=∠DCE=60°
∴∠BCD=∠ACE
∴△BCD≌△ACE（SAS）
∴BD=CE

As shown in the figure, ab = AC, ad ⊥ BC at point D, ad = AE, AB bisection ⊥ DAE intersect de at point F. please write out three pairs of congruent triangles in the graph and select one pair to prove it

(1) ADB ≌ (ADB ≌ (ADB ≌ ≌ △ ADC, △ abd AB = Acad = ad 

It is known that: as shown in the figure, ab = AC, point D is the midpoint of BC, AB bisection ∠ DAE, AE ⊥ be, and the perpendicular foot is e Confirmation: ad = AE

It is proved that: ∵ AB = AC, point D is the midpoint of BC,
∴∠ADB=90°，
∵AE⊥EB，
∴∠E=∠ADB=90°，
∵ AB bisection ∠ DAE,
∴∠1=∠2；
In △ ADB and △ AEB,
∠E＝∠ADB＝90°
∠1＝∠2
AB＝AB ，
∴△ADB≌△AEB（AAS），
∴AD=AE．

As shown in the figure, it is known that e is a point on the edge BC of diamond ABCD, and ab = AE, AE intersect BD at point O, angle DAE = 2, angle BAE

It is proved that: ? DAE = 2 ∠ BAE, ad ∥ BC  AEB = ∠ DAE
∵AB=AE∴∠ABE=∠AEB∴∠ABE=∠DAE=2∠BAE
Set options BAE=x °, so options ABE= options AEB=2x °
∴x+2x+2x=180,x=36°
∴∠ABE=∠AEB=∠DAE=72°
∴∠BAD=108°
∵ is a rhombus ᙽ AB = ad, ᙽ abd = ∠ ADB = 36 °, DOA = 72 °, △ doa is an isosceles triangle
﹣ it is necessary to prove that △ Abe and △ DOA are congruent
∠BAE=∠ADO,AB=DA,∠ABE=∠DAO
Ψ congruent
∴OA=EB