MathLabs

Problem 4

For three points A,B,CA,B,C in the plane, let m(ABC)m(ABC) be the smallest length of the three heights of triangle ABCABC, and set m(ABC)=0m(ABC)=0 when the points are collinear. Given A,B,CA,B,C, prove that for every point XX in the plane, m(ABC)≤m(ABX)+m(AXC)+m(XBC)m(ABC)\le m(ABX)+m(AXC)+m(XBC).
Step 5 of 5: Finish by the inside case
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In plain words

The outside case is reduced to the case already settled by area additivity.

m(ABD)+m(ACD)+m(BCD)≥m(ABC)m(ABD)+m(ACD)+m(BCD)\ge m(ABC)
Detailed analysis

The point D lies on BC, so the already-proved inside/boundary case applied to D gives the displayed inequality (with the degenerate triangle BCD contributing zero). Combining it with the outside reduction proves the claim for every X.