224. The Median's Secret · Why two sides always beat twice the median

Try to break it

Drag A, B, or C to reshape the triangle. AD is the median to BC; extend it to E so AE = 2·AD. Triangle ABE then satisfies AB + AC > AE = 2·AD by the triangle inequality. The median-sum inequality holds at every shape — try to find a triangle where AB + AC ≤ 2·AD; impossible.

How you build it

Construct a triangle with median extended.

• Place vertex A near the top.
• Place vertex B at the lower left.
• Place vertex C at the lower right.
• Connect A and B.
• Connect A and C.
• Connect B and C.
• Midpoint tool: click B, then C — D drops at the exact middle of BC.
• Ray tool: click A, then D. The median runs from A through D and continues past it — long enough to reach E at twice the median.
• Compass: click centre D, then click A to set the radius DA. The arc passes through A and cuts the produced median again at E, so DE = AD.
• Click where the arc crosses the ray below D to mark E. Now AE = 2·AD.
• Draw the segment from B to E. Because BDEC is a parallelogram, BE = AC.

The proof, step by step

Prove that AB + AC is greater than twice the median AD.

• AD = DE and BD = CD by construction.
• ∠ADB = ∠CDE (vertically opposite angles).
• ∴ ΔADB ≅ ΔEDC by SAS congruence.
• ∴ AB = CE (C.P.C.T.C.).
• In ΔACE, AC + CE > AE (Triangle Inequality).
• Substituting CE = AB and AE = 2AD, we get AB + AC > 2AD.

Worked example

In ΔABC, AB = 10 cm, AC = 14 cm, and AD is the median to BC. What is the possible range for the length of AD?

By the median inequality, AB + AC > 2AD ⇒ 10 + 14 > 2AD ⇒ AD < 12 cm. Also, |AB - AC| < 2AD ⇒ 4 < 2AD ⇒ AD > 2 cm. Combining these, 2 cm < AD < 12 cm.

• AD < 12 cm
• 2 cm < AD < 12 cm — correct
• AD > 12 cm
• AD = 6 cm