Why is P outside and nearer to A if λ/μ is between -1 and 0?

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matthew20103
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Does anyone know why if
(i) -1 < λ/μ < 0 then P is outside AB and nearer to A;
(ii) λ/μ < -1 then P is outside AB and nearer to B?

Can anybody give me the proof? Thanks in advance!

Additional information: AP:PB = λ:μ

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AP:PB = λ:μ is equivalent to AP/PB=λ/μ

(i) -1 < λ/μ < 0 then P is outside AB and nearer to A;


P is outside AB: |AP|>|AB|
P is nearer to A: |AP| < |BP|

-1 < λ/μ < 0
-1 < AP/PB < 0
1 > AP/BP > 0

Let AP be +ve therefore BP is +ve (since AP/BP > 0)
1 > AP/BP > 0
BP > AP > 0
BP > AP
P is nearer to A
BP > AP > 0
AB+BP > AB+AP > AB
AP > AB
P is outside AB

I can't be bothered with (ii),
 
superg33k said:
AP:PB = λ:μ is equivalent to AP/PB=λ/μ

(i) -1 < λ/μ < 0 then P is outside AB and nearer to A;


P is outside AB: |AP|>|AB|
P is nearer to A: |AP| < |BP|

-1 < λ/μ < 0
-1 < AP/PB < 0
1 > AP/BP > 0

Let AP be +ve therefore BP is +ve (since AP/BP > 0)
1 > AP/BP > 0
BP > AP > 0
BP > AP
P is nearer to A
BP > AP > 0
AB+BP > AB+AP > AB
AP > AB
P is outside AB

I can't be bothered with (ii),

Thank you brother~
But I think the condition that "P is outside AB: |AP|>|AB|" is not necessary.