As shown in the figure, in RT △ ABC, ∠ C = 90 °, a = 30 °, the bisector of ∠ C intersects with the bisector of the outer corner of ∠ B at point E, connecting AE, then ∠ AEB =?

As shown in the figure, in RT △ ABC, ∠ C = 90 °, a = 30 °, the bisector of ∠ C intersects with the bisector of the outer corner of ∠ B at point E, connecting AE, then ∠ AEB =?

From this we can see that point E is the center of gravity of the triangle, and the angle AEC can be calculated to be 120 degrees

As shown in the figure, in RT △ ABC, ad and AE are the high and middle lines on the hypotenuse BC respectively, and AF is the bisector of the angle ABC
Verification: AF is the bisector of ∠ DAF

It should be: AF is the bisector of ∠ DAE
prove:
∵ ad is the high value of △ ABC
∴∠B+∠BAD=∠B+∠C=90°
∴∠BAD=∠C
AE is the middle line
∴AE=CE
∴∠CAE=∠C
∴∠BAD=∠CAE
AF is the angular bisector
∴∠BAF=∠CAF
∴∠BAF-∠BAD=∠CAF-∠CAE
That is, DAF = EAF
Ψ AF is the bisector of ∠ DAE