Let ABC be a triangle.
H is perpendicular to AC, and intersects it on a point O such that AO=OC (O is the middle of AC).
L divides the angle B into 2 equal parts.
Then:
1. If AB=BC the H and L are on one line.
2. Otherwise, the intersection points of H and L is outside of the triangle.
Furthermore, let M be that intersection point. Let MM' be a perpendicular segment from M to BC, and MM'' a perpendicular segment from M to AB. Than one of the pair M', M'' is inside the triangle, and one is on the outside.
Proof:
1. If AB=BC, than ABC is an isoceles, and therefore H and L are one line.
2. Without loss of jenerality, assume BC>AB.
According to the Lemma of Ecker, BO divides the angle B into 2 parts, such that if BC>AB, the angle between BC and BO is smaller than the one between AB and BO.
Therefore, L is between AB and BO.
The intersection point of H and BO is O, so it's obvious the intersection point of H and L is outside of the triangle (see figure)
Now, let's assume that both M' and M'' are inside the triangle ABC (see figure).
Note that -
a. BMM'' and BMM' have equal angles and a common side BM. Therefore, BM''=BM' and MM'=MM''.
b. AOM and COM are both erect triangles. AO=OC, MO=MO, so AM=MC.
c. From a and b we got MM'=MM'' and AM=MC. AMM'' and CMM' are both erect triangles, so therefore AM''=CM'.
from a and c we got BM''=BM' and AM''=CM'. Therefore, BM''+AM''=BM'+CM'.
We got AB=BC, which contradicts the assumption that BC>AB.
In a similar way we prove that M' and M'' can't be both outside of the triangle ABC, and that concludes the theorem.
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