Proof Revision!

Geometry Level 3

Take a look at the following proof.

We're going to try to prove that if $O$ is a point inside $\triangle ABD$ , then $AD+AB>OD+OB$ .

First join $O$ and $A$ .

From the triangle inequality,

$AD+OA> OD\cdots (1)$

$OB+OA>AB\cdots (2)$

Subtract $(2)$ from $(1)$ to get,

$AD-OB>OD-AB$

Switch sides to get,

$AD+AB>OD+OB$

$\mathbb{QED}$

Here are some comments about the proof.

$[1]$ . The proof is not correct. There's an invalid move hiding in there somewhere.

$[2]$ . Relax! Not everything in this set is a trick question. This proof is perfectly fine.

$[3]$ . Forget the proof! The claim itself is not true!

Which comment here is correct?

This problem is from the set "MCQ Is Not As Easy As 1-2-3". You can see the rest of the problems here .

[2]

None of them are correct. You missed the letter

C

in the diagram.That's why this proof is wrong!

[3]

[1]

1 solution

Mursalin Habib
May 12, 2014

The proof is actually invalid. You can't just subtract inequalities like that. Watch what happens when you subtract $(1)$ from $(2)$ .

You get,

$OB-AD>AB-OD$

Rearrange the terms to get,

$OD+OB>AD+AB$ .

This is exactly the opposite of what you have to prove. Inequalities can not be subtracted that way. We just happened to get a correct conclusion from an incorrect method.

If $a>b$ and $c>d$ , the correct way to 'subtract' them is multiplying the latter by $-1$ and then adding them together.

So, the $c>d$ becomes $-d>-c$ and $a-d>b-c$ . This last statement is always correct.

As an example, take these two inequalities

$6>5\cdots (1)$

$5>3\cdots (2)$

Subtract $(2)$ from $(1)$ and you're going to get

$1>2$

which is obviously not true.

Here is a correct proof that $AD + AB > OD + OB$ :

Let $C$ be the intersection of $AB$ and $OD$ . By the triangle inequality on triangle $BCO$ , $BC + CO > BO.$ By the triangle inequality on triangle $ACD$ , $AC + AD > CD.$ We can certainly add these inequalities, to get $AC + BC + AD + CO > BO + CD = BO + CO + OD.$ We can write $AC + BC = AB$ , and cancel $CO$ from both sides, to get $AB + AD > BO + OD.$

Jon Haussmann - 7 years, 1 month ago

This eliminates statement $[3]$ as we can see that the claim is of course true.

Mursalin Habib - 7 years, 1 month ago

Yes, the fist statement in the proof the subtraction of inequality is wrong ,First upvoter

Mardokay Mosazghi - 7 years, 1 month ago

This question reminds me of two questions I came up with during a test :)

Tan Li Xuan - 7 years, 1 month ago

Statement is correct and though I knew you subtracted inequalities, I perchance clicked the on wrong one....my luck ;)

kalyan pakala - 7 years ago

Proof Revision!

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