220. The Difference Rule · Two sides can't differ by more than the third

Try to break it

Drag any vertex. For every position, |AB − BC| < AC, |BC − CA| < AB, and |CA − AB| < BC — the difference of any two sides is always less than the third. Try to find a position where this fails; if it ever did, the sides couldn't close into a triangle.

How you build it

Construct a triangle from three sides and test the triangle inequality.

• Point tool: mark point Q.
• Point tool: mark point R — QR is the first (base) side.
• Segment tool: join Q to R.
• Arc tool: click centre Q and drag out to the length you want for side PQ.
• Arc tool: click centre R and drag out to the length you want for side PR. The two arcs cross at P — but only if each side is longer than the difference of the other two.
• Point tool: mark P where the two arcs cross.
• Segment tool: join P to Q.
• Segment tool: join P to R. A triangle forms only because each side exceeds the difference of the other two.

The proof, step by step

Prove that the difference of any two sides of a triangle is less than the third side.

• In any triangle, the sum of the lengths of any two sides is greater than the length of the third side.
• Consider sides AB, BC, and AC. We know AB + BC > AC.
• Subtract BC from both sides: AB > AC - BC.
• This means the difference AC - BC is strictly less than AB.
• The same logic applies to the other two pairs of sides, proving the corollary.

Worked example

In triangle PQR, PQ = 12 cm and QR = 8 cm. Which of the following could be the length of side PR?

By the corollary, |PQ - QR| < PR < PQ + QR. So, |12 - 8| < PR < 12 + 8, which means 4 < PR < 20. Only 5 cm lies strictly between 4 and 20.

• 3 cm
• 4 cm
• 5 cm — correct
• 20 cm