Triangle inequality theorem proof
Before you understand the triangle inequality theorem proof, you need to review the
triangle inequality theorem and understand the shortest distance theorem.
Shortest distance theorem:
The shortest distance from a point p to a line s is the line perpendicular to s and passing through p
This is illustrated below. As you can see the shortest distance is segment PR and this is shown in blue.
Any other segments such as segment PF or segment PO ( shown in red) is longer.
You have to understand this theorem before trying to comprehend the proof about triangle inequality theorem
Now here is ther proof:
Draw any triangle ABC and the line perpendicular to BC passing through vertex A. (This is shown in blue)
Now prove that BA + AC > BC
BE is the shortest distance from vertex B to AE
This means that BA > BE
By the same token,
CE is the shortest distance from C to AE
This means that AC > CE
Let us put it all together:
BA > BE and
AC > CE
Add the left side and add the right of the inequalities. This gives:
BA + AC > BE + CE
Now, notice that BE + CE = BC
Therefore, BA + AC > BC
Now, starting with the same triangle, draw the line perpendicular to AC passing through vertex B. (This is shown in blue)
Prove that BA + BC > AC
AE is the shortest distance from vertex A to BE
This means that BA > AE
By the same token,
CE is the shortest distance from C to BE
This means that BC > CE
Let us put it all together:
BA > AE and
BC > CE
Add the left side and add the right of the inequalities. This gives:
BA + BC > AE + CE
Now, notice that AE + CE = AC
Therefore, BA + BC > AC
Now, here is your exercise: Try to prove that AC + BC > AB

Nov 09, 18 09:40 AM
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