Proof of triangle inequality theorem

Before you understand the triangle inequality theorem proof, you must consider the triangle inequality theorem and understand the shortest distance theorem.

Shorter distance theorem:

The shortest distance of the point s p in a line is the line perpendicular to the s and passing by p

This is illustrated below. As you can see, the shortest distance is the PR segment, and this is shown in blue.

Triangle-inequality-illustration-imageAll other segments such as the segment PF or segment PO (indicated in red) is longer.

You must understand this theorem before trying to understand the evidence on the triangle inequality theorem

Now, here are the other evidence:

Draw any triangle ABC, and the line perpendicular to BC passing by vertex a. (this is shown in blue)

Triangle-inequality-illustration-imageNow prove that BA + AC > BC

BE is the shortest distance from vertex b in AE

This means that BA > BE

At the same time.

This is the shortest distance between c and AE

This means that AC > This

Let us put all together:

BA > BE and

AC > THIS

Add the left and the right of the inequality. This gives:

BA + AC > BE + IT

Now, note that BE + CE = BC

Therefore, BA + AC > BC

Now, starting with the same triangle, draw a line perpendicular to the CA via vertex (this is shown in blue) b.

Triangle-inequality-illustration-imageProve that BA + BC > AC

AE is the short distance from vertex has to BE

This means that BA > AE

At the same time.

This is the shortest distance from c to BE

This means that BC > This

Let us put all together:

BA > AE and

BC > THIS

Add the left and the right of the inequality. This gives:

BA + BC > AE + EC

Now, note that AE + CE = CA

Therefore, BA + BC > AC

Now, here is your exercise: try to prove that AC + BC > AB

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