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 4 of 5: Reduce an outside point
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In plain words

Moving an outside point back to the opposite side cannot decrease the relevant minimum heights.

m(ABX)+m(ACX)≥m(ABD)+m(ACD)m(ABX)+m(ACX)\ge m(ABD)+m(ACD)
Detailed analysis

For X outside, take A to be the farthest of A,B,C from X. If ABCX is concave, one vertex (say B) lies in triangle ACX; comparing the relevant rays gives m(ACX) at least m(ABC). If ABCX is convex, let D=AX intersection BC. A case check according to which vertex supplies the minimum altitude shows m(ABX) at least m(ABD), and analogously m(ACX) at least m(ACD).