In the triangle ABC, the points D and E are on BC, the angle CAE is equal to angle B, e is the midpoint of DC, BD is equal to AC, and the ad bisection angle BAE (1) when the angle BAC is equal to 90 degrees In the triangle ABC, points D and E are on BC, the angle CAE is equal to angle B, e is the midpoint of DC, BD is equal to AC, and ad bisects the angle BAE (1) When the angle BAC is equal to 90 degrees, it is proved that BD is equal to AC (2) When the angle BAC is not equal to 90 degrees, is there still BD equal to AC? Explain the reason

From CE = ed, AE = EF, △ ade ≌ △ FCE (s, a, s). Therefore, BD = AC is obtained
DA=CF,

As shown in the figure, in △ ABC, points D and E are on edge BC, ∠ CAE = ∠ B, e is the midpoint of CD, and ad bisection ∠ BAE proves that BD = AC

Make an auxiliary line and make an extension line along AE to f so that AE = ef (i.e., e is the midpoint of AF). Connect BF
It is proved that since e is the midpoint of CD and AF, ADFC is a parallelogram, so ∠ AFD = ∠ CAE, AC = DF
Because ad bisects ∠ BAE, so ∠ bad = ∠ fad
Because △ abd is congruent with △ AFD, so AB = AF, so ABF = ∠ AFB
Because AFD = ∠ B, so ∠ DBF = ∠ DFB, so BD = DF
AC = DF, so BD = AC

As shown in the figure, it is known that in the triangle ABC, the angle cab = 90 degrees, the point D and E are on the edge BC, the angle CAE = the angle B, e is the midpoint of CD, ad bisects the angle BAE, verification: BD = AC

Angle CAE = angle B
Angle cab = 90 degrees
AE vertical BC and E are obtained,
There are
E is the midpoint of CD
The angle CAE = angle EAD
There are
Ad bisection angle BAE
The angle CAE= angle EAD= angle DAB=30 degrees is obtained
AC = DC = BD
AC = BD

In the right angle △ ABC, ∠ BAC = 90 °, points D and E are on edge BC, ∠ CAE = ∠ B, e is the midpoint of CD, and ad bisects ∠ BAE Question: is BD equal to AC? Please tell me your reason

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As shown in the figure, △ ABC, ab = AC, ad = De, ∠ bad = 20 ° and ∠ EDC = 10 °, then the degree of ∠ DAE is () A. 30° B. 40° C. 60° D. 80°

Let ∠ C = x, ∵ AB = AC
∴∠B=∠C=x
∴∠AED=x+10°
∵AD=DE，∴∠DAE=∠AED=x+10°
According to the theorem of X + 10 ° of the triangle, x + 10 ° is obtained
If x = 50 °, then ∠ DAE = 60 °
Therefore, C

19. Given △ ABC, make isosceles △ abd and isosceles △ ace so that ab = ad, AC = AE, ∠ bad = ∠ CAE, the straight lines CD and be intersect at point o (1) As shown in Figure 1, if ∠ abd = ∠ CAE = 60 °, then ∠ BOD= (2) As shown in Figure 2, if ∠ abd = ∠ CAE = a, a is an acute angle, then ∠ BOD= (3) As shown in Fig Question two

(3) In the abod of quadrilateral, the value of BOD + bad + ABO + ADO is 360
Also, ∠ ABO = ∠ ADC
∠ADC+∠ADO=180°,
Therefore, ∠ ADO + ∠ ABO = 180 °,
Therefore, ∠ BOD + ∠ bad = 180 °
Then, a + ∠ BOD = 180 ℃
The first and second questions can be brought in

In triangle abd, ab = ad, in triangle ace, AC = AE, point C is on edge BC, and BC = De, prove that angle bad = angle CAE

prove:
∵AB=AD,AC=AE,BC=DE
∴⊿ABC≌⊿ADE（SSS）
∴∠BAC=∠DAE
∴∠BAC+∠CAD=∠DAC+∠CAD
That is ∠ bad = ∠ CAE

In this paper, we present a new method for solving the problem of ad-b (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

In △ abd and △ ace, ab= AD.AC=AE The intersection of BC and de at f.bc and ad at point G If DF 2 = FG · FB., then BC bisects ∠ abd. Why?

Because of [BAD=] [CAE], [BAD+] [DAC=] [CAE+] [DAC], that is, [BAC=] [DAE]
Because: ab = ad, AC = AE
So the triangle DAE is all equal to the triangle ABC
So: ∠ ABC = ∠ ade
And because: DF 2 = FG · FB, so: DF / FG = FB / DF
In triangle DFG and triangle BFD, ∠ DFG = ∠ BFD, DF / FG = FB / DF
So triangle DFG is similar to triangle BFD
So ∠ GDF (∠ ADE) = ∠ DBC
So we have: ∠ ABC = ∠ DBC
So: FB bisection ∠ abd

As shown in the figure, the isosceles right triangle △ abd and △ ace are respectively made with the sides AB and AC of △ ABC as the right angles. It is proved that the angle AFD = the angle AFE

The estimated point F should be the intersection point of be and DC. It is proved that if DAB = CAE = 90 ° then DAC = ∠ BAE; (equal plus equal sum, etc.) and ad = AB; AC = AE. So ⊿ DAC ≌ Δ BAE (SAS), be = DC