The triangle ABC is an equilateral triangle, the triangle ABC is an isosceles triangle, ad = AE, ∠ DAE = 80 °, bad = 15 ° to find the degrees of ∠ CAE and ∠ EDC

Ade is isosceles triangle, the vertex angle is 80, so the angle ADF = 50
Angle bad = 15, so BDA = 105
Angle FDC = 180 - angle ADF - angle BDA = 25
There is no chart comparison

As shown in the figure, in the triangle ABC, it is known that ad is equal to AE, AB is equal to be and CD is equal to ac. it is proved that △ abd is congruent △ ace and ∠ bad are equal to ∠ CAE

∵AB=BE=CD=AC
∴AB=AC
Be = CD, namely BD + de = de + CE
∴BD=CE
In △ abd and △ ace:
∵AD=AE,AB=AC,BD=CE
∴△ABD≌△ACE(SSS)
∴∠BAD=∠CAE

It is known that in △ ABC, ∠ BAC = 45 ° with AB and AC as the edges, isosceles △ abd and △ ace are made outside △ ABC, ab = ad, AC = AE, and ∠ bad = CAE, connected with CD Be and F, AF (1) (1) as shown in Figure 1, if ∠ bad = 60 °, then ∠ AFE= ② As shown in Figure 2, if ∠ bad = 90 °, then ∠ AFE= ③ As shown in Figure 3, if ∠ bad = 20 °, then ∠ AFE= (2) As shown in Fig. 4, if ∠ bad = a °, guess the degree of ∠ AFE and prove it (3) As shown in Fig. 5, the degree of ∠ AFE is written directly by △ abd in Fig. 2 clockwise around point a (45 ° < B < 90 °)

(1) If ∠ bad = 60 °, then ∠ AFE = 60 degrees
If bad = 90 °, AFE = 45 degrees
If bad = 20 °, AFE = 80 degrees
(2) Angle AFE = 90 degrees - 1 / 2A
As shown in the figure, AP and AQ are perpendicular to DC and be respectively through a, and the perpendicular feet are p and Q. let be and AC intersect with point o
It can be proved that triangle DAC and BAE are congruent
So the angle AEB = ACD
Because the angle AOE = FOC
So the angle CFE = CAE = a
Because the angle AEB = ACD, AQE = APC = 90 degrees, AE = AC
So the triangle AQE and APC are congruent
So AP = AQ, that is, the distance from point a to FD, Fe is equal
So FA bisection angle DFE
So the angle AFE = 1 / 2dfe = 1 / 2 (180 degrees - EFC) = 1 / 2 (180 degrees - a) = 90 degrees - 1 / 2A
(3) Angle AFE = 135 degrees
As shown in the figure, AP '= AQ'
So point a is on the bisector of angle b'f'c, that is, f'a bisector angle b'f'c
Angle b'f'c = f'b'd '+ F'd'b' = AB'd '+ ad'b' = 90 degrees
So the angle af'c is 45 degrees
So the angle af'e '= 135 degrees

As shown in the figure, in △ abd and △ ade, ab = ad, AC = AE, ∠ bad = ∠ CAE, connecting BC and de intersecting at point F, BC and ad intersecting at point G (1) Try to judge the quantitative relationship between line BC and De, and explain the reasons; (2) If ∠ ABC = ∠ CBD, is segment FD the proportional median of segments FG and FB? Why?

(1) The quantitative relationship between BC and De is BC = de. the reasons are as follows: ∵ bad = ∠ CAE,  bad + ∠ DAC = ∠ CAE + ∠ DAC, that is ∵ BAC = ∠ DAE, and ∵ AB = ad, AC = AE, ≌≌△ ade. (SAS) ∵ BC = de. (2) segment FD is the middle term of the proportion of FG and FB

As shown in the figure, in △ abd and △ ace, ab = ad, AC = AE, ∠ bad = ∠ CAE, connecting BC and de intersecting at point F, BC and ad intersecting at point g. verification: BC = De

Proof: ∵ bad = ∵ CAE,
∴∠BAD+∠DAC=∠CAE+∠DAC，
In other words, ∠ BAC = ∠ DAE
In △ cab and △ EAD
AB＝AD
∠BAC＝∠DAE
AC＝AE ，
∴△CAB≌△EAD（SAS），
∴BC=DE．

It is known that △ abd and △ ace are isosceles, and ∠ bad and ∠ CAE are right angles. It is proved that △ ACD is equal to △ AEB. The relationship between ∠ AFD and ∠ AFE is judged and proved

∠AFD=∠AFE
prove:
∵∠CAE=∠BAD=90°
∴∠CAD=∠BAE
∵AD=AB,AC=AE
∴△ADC≌△ABE（SAS）
∴CD=BE
The area of △ ACD = △ Abe
The distance from point a to CD = the distance from point a to be (the area is equal, the bottom is equal, so the height is equal)
ν A is on the bisector of ∠ DFE
∴∠AFD=∠AFE

In △ abd and △ ace, ∠ BAC = ∠ CAE = 90 °, ad = AB, AC = AE, proving ∠ AFD = ∠ AFE, sorry, no figure

There is no f-point in the known. Why does f-point suddenly appear in the conclusion of proof? In addition, without graph, where does f-point come from?

As shown in the figure, in △ abd and △ ade, ab = ad, AC = AE, ∠ bad = ∠ CAE, connecting BC and de intersecting at point F, BC and ad intersecting at point G (1) Try to judge the quantitative relationship between line BC and De, and explain the reasons; (2) If ∠ ABC = ∠ CBD, is segment FD the proportional median of segments FG and FB? Why?

Given △ ABC, we make △ abd and △ ace with AB and AC as edges respectively, and ad = AB, AC = AE, ∠ DAB = ∠ CAE, connect DC and be, G and F are DC and be respectively (1) if ∠ DAB = 60 °, the degree of ∠ AFG is (2) if DAB = 90 ° then the degree of ∠ AFG is____ (3) if ∠ DAB = x °, try to explore the quantitative relationship between ∠ AFG and X and give proof

1. If ∠ DAB = 60 °, the degree of ∠ AFG is 60 °
2. If ∠ DAB = 90 °, the degree of ∠ AFG is 45 °
3. If ∠ DAB = x °, then the quantitative relationship between ∠ AFG and X is as follows:
∠AFG =（180°- x°）/2
The proof is as follows:
∵∠DAB =∠CAE
∴∠DAC =∠BAE
Ad = AB, AC = AE
∴△DAC ≌△BAE（SAS）
∴DC = BE,∠ADC =∠ABE
And ∵ g and F are the midpoint of DC and be respectively, then DG = BF
∴△DAG ≌△BAF（SAS）
∴AG = AF
If Ag = AF, then △ AFG is an isosceles triangle
∴∠AFG =（180°-∠FAG ）/2
Taking a as the rotation center, the △ DAG and △ BAF are overlapped and restored,
Visible ∠ fag = ∠ DAB
∴∠AFG =（180°-∠DAB）/2

In eighth grade mathematics, it is known that the triangle ABC takes AB and AC as sides to make triangle abd and triangle ace respectively, and ad = AB, AC = AE, angle DAB = CAE, connecting DC and be g. F is the midpoint of DC and be respectively

(1) Connecting Ag because ∠ DAB = ∠ CAE, and ∠ DAC = ∠ DAB + ∠ BAC, ∠ BAE = ∠ CAE + ∠ BAC, so ∠ DAC = ∠ BAE and ad = AB, AC = AE, know △ DAC and △ BAE are congruent, so DC = be, ∠ DCA = ∠ bea and G, f are the midpoint of DC and be respectively. It is easy to prove that triangular AGC and triangular AFE are all equal, so Ag = AF, ∠ CAG = ∠ EAF is from

The triangle ABC is an equilateral triangle, the triangle ABC is an isosceles triangle, ad = AE, ∠ DAE = 80 °, bad = 15 ° to find the degrees of ∠ CAE and ∠ EDC