Angle between a bisector of angle of a triangle and a perpendicular drawn from the same vertex to it's opposite side.

Hello math big(🐞) & hivers(🐝)
I hope you are strong and stout and doing well in life.

Well come back to another interesting geometric problem and it's solution with proof.

✅The angle between a perpendicular drawn from a vertex of a triangle and bisector of the angle has a relation to the other two angle of the triangle.You
can check the problem in the following figure.

✅Let's go to the proof of the problem.We need to find ∠BAE using angle sum property of triangle in term of b and θ. So ∠BAE=(90°-b+θ) and again using same property we find ∠EAC=(90°-c-θ).∠BAE and *∠EAC" are equal as AE is the bisector of ∠BAC.So, The solution is is given by as following:

✅The solution is quite difficult when values of b and c are not given.How will the problem look when b and c are given as shown below:
✅If it is given, you can easily find out. Using different property of triangle like angle sum property, exterior and interior angle property etc.I have also added it for better understanding. Check it below:

✅You have seen the solution of the problem when the triangle is an acute angle triangle but how it would be in case of a obtuse angle triangle. Yes the perpendicular will be on the extended opposite side of the particular vertex of the triangle. Check it below:
✅It doesn't matter what kind of triangle it is the answer is always given by half of (b-c)

✅For your curiosity let's check how it would be when the triangle is a right angle triangle🤣🤣Check the solution in the figure below:

POINT TO BE REMEMBERED:

This will work when you draw a perpendicular from any vertex and the bisector of the angle at same vertex.Do not try thinking it for median.To draw this solution Medians comes no where.I too used to do the mistake in early stage.🤣🤣

I hope you have liked my today's work. sorry not a day's work it is.

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Have a good day.

All is well.

Regards: @meta007