In rectangular ABCD, AE is perpendicular to BD, and the angle DAE = 2 angle BAE?

In rectangular ABCD, AE is perpendicular to BD, and the angle DAE = 2 angle BAE?

Solution: a quadrilateral ABCD is a rectangle
∴AO=OB,∠BAD=∠DAE+∠BAE=90
∵∠DAE＝2∠BAE
∴∠BAE=30
∵AE⊥BD
∴∠AEB=90
∴∠ABC=90-30=60
∵AO=BO
The ∧ ABO is an equilateral triangle
∴∠BAO=60
∵AE⊥BD
The AE is the high line of edge Bo
AE is also the angular bisector of Bao
∴∠EAO=∠BAO-∠BAE=60-30=30

In rectangular ABCD, AE is perpendicular to BD, and the angle DAE is 4 to 1 than the angle BAE. The degree of angle EAO is calculated

Let ∠ BAE = x, ∠ DAE = 4x, then
x+4x=90
x=18°
∠ABD=90-18=72°
∴∠EAO=72-18=54°

In rectangular ABCD, AE is vertical BD, perpendicular foot is point E, angle DAE = 3-angle BAE, then angle CAE =? Sharp
I want to know why?

Angle CAE = angle BAE = 15 degrees

As shown in the figure, in the rectangular ABCD, AE is vertical to BD, the ratio of the degree of E, DAE and EAB is 3:2, and then calculate CAE

Connect AC and BD to o
∵ rectangle ABCD
∴∠DAB＝90°,∠CAD＝∠BDA
∵∠ DAB = ∠ DAE + ∠ EAB, and the ratio of degree of ∠ DAE and ∠ EAB is 3:2
∴∠DAE＝90°*3/（2+3）＝54°
∠EAB＝90°*2/（2+3）＝36°
∵AE⊥BD,∠ABD＝∠EBA
Ψ△ DBA is similar to △ Abe
∴∠BDA＝∠EAB＝36°
∴∠CAD＝36°°
∴∠CAE＝∠DAE-∠CAD＝54°-36°＝18°

In the triangle ABC, the bisectors of ∠ ABC and ∠ ACD intersect at point A1, and the bisectors of ∠ a1bc and ∠ a1cd intersect at point A2 And so on, the bisectors of ∠ ANBC and ∠ ancd intersect at the point an. (1) find the degree of ∠ A1; (2) directly write the degree of ∠ an. (expressed by an equation containing n),

1) Because of ∠ a1cd = 1 / 2 ∠ ACD = 1 / 2 (∠ BAC + ∠ ABC) = ∠ a1bc + 30, so ∠ A1 = 30
2) According to the method of 1), it can be seen that: ∠ an = 60 / 2 ^ n

In triangle ABC, D is the point on the extension of BC, and the bisector of angle ABC and angle ACD intersects point A1
(1) Try to judge the relationship between angle A and angle A1. (2) if angle a = 96 ', the bisector of angle a1bc and angle a1cd intersects at point A2, and so on, find angle 5
The quantitative relationship between angle A and angle A1 is also judged and proved

(1)∠A1＝1/2∠A
Reason: A1 = a1cd - a1bc
∠A＝∠ACD－∠ABC＝2（∠A1CD－∠A1BC）＝2∠A1
(2) Similarly, ∠ a = 2 ＾ 5 (∠ A1)

In the triangle ABC, angle a = 96 degrees, extend BC to D, and the bisector of angle ABC and angle ACD intersects at point A1,
The bisector of angle a1bc and angle a1cd intersects at point A2, and so on, the bisector of angle ANBC and angle ancd intersects at point an + 1
Question:
1. Conjecture the formula of degree of angle an (n is a positive integer)
2. Use your conjectured formula to find the degree of angle an

How many times? Can you do it?

As shown in the figure, in △ ABC, ab = AC, De is the vertical bisector of AB, the perpendicular foot is D, the intersection of AC and E, known as ∠ Abe = 40 °, find the degree of ∠ EBC

∵ De is the vertical bisector of AB, ∵ AE = be, ∵ a = ∵ Abe, ∵ Abe = 40 °, ∵ a = 40 °, ∵ AB = AC, ∵ ABC = ∵ C = 12 (180 ° - a) = 70 °, ∵ EBC = ∵ ABC -