I got the correct answer and was able to find the solution but I have a general doubt about perpendicular drawn from the vertex of the triangle to the BC (in this case). I used to think that if the perpendicular is drawn from one of the vertices of the triangle to the base (here BC) would always bisect the angle A in this case because if the perpendicular cannot be tilted etc so how come in this question x has a different measure and A-x has a different measure. Is that not true that perpendicular has to bisect the vertex angle? In what cases it will bisect the angle?
I think you’re just confused cuz that doesn’t make much sense. Let the foot of the perpendicular be N then is \triangle CAN \cong \triangle BAN?
If your answer was “yes” then the answer would’ve been that the altitude (perpendicular from the vertex) is also an angle bisector. However, if it isn’t then why should the perpendicular also be an angle bisector? The idea of “tilt” makes little sense to consider here.
As a bonus, you can think about how this relates to altitudes of an equilateral, isosceles, and scalene triangles. When are said altitudes also angle bisector and why…?
That makes sense. I confused it with equilateral and isosceles triangles case because they have bisectors due to symmetry from the vertex but every triangle need not have that. Thank you for clarifying. Appreciate this!
