As shown in the figure, D is on the BC side of △ ABC, AC and de intersect at point F, ab = ad, ∠ bad = ∠ EAC ∠ EDC

As shown in the figure, D is on the BC side of △ ABC, AC and de intersect at point F, ab = ad, ∠ bad = ∠ EAC ∠ EDC

There is a mistake in the condition of this question. What our teacher said is that we should change the condition "bad = ∠ EAC - ∠ EDC" to "bad = ∠ EAC = ∠ EDC", and then prove it again, as follows
Proof: because ∠ bad = ∠ EAC
So ∠ bad + ∠ DAC = ∠ EAC + ∠ DAC
That is, BAC = DAE
Because ∠ EAC = ∠ EDC, ∠ AFE = ∠ DFC
So 180 - ∠ EAC - ∠ AFE = 180 - ∠ EDC - ∠ DFC
That is, AED = ACD
In △ ABC and △ ade
∠BAC = ∠DAE
∠AED=∠ACD
AB=AD
So △ ABC ≌ ade (AAS)
So BC = de

In an isosceles triangle, ab = AC, C, D is the triad point of BC, and find ∠ bad = ∠ DAE = ∠ EAC

BD = (1 / 2) AB; known, 2CD = 3AB, we can get: CD = (3 / 2) AB; known, e, f is the trisection of CD, we can get: DF = (1 / 3) CD = (1 / 2) AB = ad = BD; then: △ ADF and △ BDF are isosceles right triangle, we can get: ∠ AFD = ∠ BFD = 45 °; so, ∠ AFB = ∠ AFD

Points D and E are on the edge BC of triangle ABC. If BD = ad = AE = EC and ∠ B = 2 ∠ DAE, then ∠ BAC=

Let ∠ DAE = x, then ∠ B = 2x, because: BD = ad, so ∠ DAB = ∠ B = 2x ∠ ade = ∠ DAB + ∠ B = 4x, because: ad = AE, so ∠ AED = ∠ ade = 4x, because: AE = EC, so ∠ EAC = ∠ C = 1 / 2 ∠ AED = 2x, so: ∠ BAC = ∠ bad + ∠ DAE + ∠ EAC = Z2 + X + 2x = 5x, because ∠ BAC + ∠ B + ∠ C = 5x + 2x + 2x = 9

As shown in the figure, in △ ABC, D and E are two points on the edge of BC, and BD = de = EC, then ad is the middle line of the triangle and AE is the middle line of the triangle

Ad is the middle line of triangle Abe, AE is the middle line of triangle ADC