Let us see the scheme at work on a complete problem, following the six steps one by one.

Example — Applying the scheme

Problem: In the triangle ABCABC, let MM be the midpoint of BCBC. Prove that AM<AB+AC2AM < \dfrac{AB+AC}{2}.

1. Figure: triangle ABCABC with MM the midpoint of BCBC. I extend AMAM by a segment MDAMMD\cong AM.

2. Hyp: BMMCBM\cong MC.    Thes: AM<AB+AC2AM < \dfrac{AB+AC}{2}.

3. Backward Thinking: it is enough to show 2AM<AB+AC2\,AM < AB+AC, that is AM+MD<AB+ACAM+MD < AB+AC, that is AD<AB+ACAD < AB+AC.

4. Construction: I extend AMAM beyond MM by MD=AMMD = AM and I join DD to CC.

5. Triangles: AMB\triangle AMB and DMC\triangle DMC have AMMDAM\cong MD, BMMCBM\cong MC, AMB^DMC^\widehat{AMB}\cong\widehat{DMC} (vertically opposite)     \implies by the first criterion     \implies ABDCAB\cong DC.

6. Conclusion: By the triangle inequality in ADC\triangle ADC: AD<AC+DC=AC+ABAD < AC + DC = AC + AB. Therefore 2AM=AD<AB+AC2\,AM = AD < AB + AC, that is AM<AB+AC2AM < \dfrac{AB+AC}{2}. \blacksquare

Topics: Euclidean geometry
Concepts: Congruence criteria · Proof · Triangle inequality · Median
Skills: Proving · Synthetic geometry