MathLabs

Problem 1

Let ABCABC be a triangle with incenter II. A point PP in the interior of the triangle satisfies ∠PBA+∠PCA=∠PBC+∠PCB\angle PBA+\angle PCA = \angle PBC+\angle PCB. Show that AP≥AIAP \geq AI, and that equality holds if and only if P=IP=I.
Step 4 of 4: Apply the triangle inequality to A,P,MA,P,M
AP+PM≥AM=AI+IM ⟹ AP≥AI(equality iff P=I)AP+PM\ge AM=AI+IM\ \Longrightarrow\ AP\ge AI\quad(\text{equality iff }P=I)
Detailed analysis

By the triangle inequality on A,P,MA,P,M, we have AP+PM≥AM=AI+IMAP+PM\ge AM=AI+IM. Subtracting PM=IMPM=IM (Step 3) gives AP≥AIAP\ge AI. Equality holds in AP+PM≥AMAP+PM\ge AM if and only if PP lies on segment AMAM; since segment AMAM meets the circle centered at MM of radius MIMI only at the single point II, equality holds if and only if P=IP=I.