As shown in the figure, in △ ABC, AE bisects ∠ BAC, ∠ C ＞ B, f is the point above AE, and FD ⊥ BC is at point D. try to deduce the relationship between ∠ EFD, ∠ B and ∠ C When the point F is on the extension line of AE, as shown in Figure 2, the other conditions remain unchanged. Is the conclusion you deduced in question 1 still valid? Please give reasons.

As shown in the figure, in △ ABC, AE bisects ∠ BAC, ∠ C ＞ B, f is the point above AE, and FD ⊥ BC is at point D. try to deduce the relationship between ∠ EFD, ∠ B and ∠ C When the point F is on the extension line of AE, as shown in Figure 2, the other conditions remain unchanged. Is the conclusion you deduced in question 1 still valid? Please give reasons.

Proof: Ag ⊥ BC
Known FD ⊥ BC FD / / Ag
∠ EFD = ∠ EAG (both cases are the same)
Let ∠ EAG = ∠ EFD = a
∠CAG=x
Then ∠ C = 90 ° - x EA bisects ∠ a
∠BAE=∠CAE=∠EAG+∠CAG=X+a
∠BAG=∠BAE+∠EAG=X+2a
∴∠B=90°-X-2a ∠C=90°-X
a=（（90°-X）-(90°-X-2a)）/2
∠EFD=(∠C-∠B)/2

In the triangle ABC, G is the center of gravity, passing through G for EF / / BC, intersecting AB at point F, then s triangle AEF:S Triangle ABC=_______

Because of the triangle, the center of gravity of the triangle is the intersection of the middle line of the triangle
It can be proved that the distance from the center of gravity to the vertex is twice the distance from it to the midpoint of the opposite side
So let the extension of Ag intersect BC at h, so Hg: Ag = 1:2, then Ag: ah = 2:3
So the s triangle AFE:S Triangle ABC = 4:9

It is known that the three sides of △ ABC and △ def satisfy AB ／ de = AC ／ DF = BC ／ EF, and ∠ a = 56 °, B = 84 °, then ∠ d =, ∠ f =

It is known that the three sides of △ ABC and △ def satisfy AB ／ de = AC ／ DF = BC ／ EF
So △ ABC is similar to △ def
And ∠ a = D, ∠ B = e, ∠ C = F
It is also known that a = 56 ° and B = 84
Therefore, d = 56 ° and F = 180-56-84 = 38