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In the triangle ABC, AE bisects the angle BAC (angle c is greater than angle b), f is the point above AE, FD is perpendicular to BC and D. try to deduce the quantitative relationship between angle EFD and angle B, angle C
I think it can be divided into two cases. First, f is the point on AE, and it does not say whether it is the extension line or the AE line segment
(1) When f is a point on line AE: 1. AE bisects ∠ BAC
Therefore, BAE = EAC
In EFD of right triangle
∠FED=∠B+∠BAE=180°-∠BEA=180°-∠C-∠EAC
∠EFD=90°-∠FED=90°-(∠B+∠BAE)=90-(180°-∠C-∠EAC)=∠C+∠EAC-90°
The equation is obtained
①∠EFD=90-∠B-∠BAE
②∠EFD=∠C+∠EAC-90=∠C+∠BAE-90°
①+②=
2∠EFD=∠C-∠B
Therefore, the relationship between ∠ EFD and ∠ C ∠ B is significant
∠EFD=1/2(∠C-∠B)
(2) When f is on the AE extension line
2. Through point a, Ag ⊥ BC is set at G
It is known from (1) that ∠ EAG = (∠ C - b)
∵ AG⊥BC,∴∠AGB=90°,
∵ DF⊥BC,∴∠FDC=90°,
∴∠AGB=∠FDC,∴ FD‖AG ．
∴∠AFD=∠EAG．
∴∠AFD=（∠C－∠B）．
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