Let's assume a right triangle where ∠BAC =90° and a perpendicular has been drawn to hypotenuse "BC" from point "A". And this perpendicular intersects hypotenuse at point "D" such that ∠ADB and ∠ADC equal to 90°. Does this perpendicular bisects the angle ∠BAC? One video lecture says that this perpendicular does not bisect ∠BAC, i.e ∠BAD and ∠CAD are not equal, and they are not 45°. The link of this video lecture is given below:
Let's assume another figure where an equilateral triangle has been drawn inside a circle. Each angle Angle of that equilateral triangle is 60°. Central angle∠AOC drawn on the arc AC is twice the inscribed angle ∠ABC drawn on that same arch AC. Since ∠ABC equals to 60°, ∠AOC is 120°. A perpendicular has been drawn to line AC, bisecting the angle ∠AOC into two equal half, i.e ∠AOD and ∠COD are equal to 60°. And OD is perpendicular to AC. (These lines about equilateral triangle drawn inside a circle have been taken from a video lecture made by Khan Academy link of which has been given below).
In case of 1, perpendicular drawn from point A to hypotenuse "BC" doesn't bisect the angle BAC, and in case of 2, perpendicular drawn from point "O" to line AC bisects ∠AOC. Please, let me know when a perpendicular drawn through an angle to a line of a triangle bisects that angle and when it doesn't.