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For triangle ABC, there is a point X such that B-A-X. There is any point P on bisector of exterior angle CAX. Prove that PB+PC>AB+AC.

I have tried for many hours, but i have no idea how to solve it. actually, i have shown PB>AB. This is a question in 9th grade level, so please give solution by use of triangle congruency and triangle inequality.. Please don't use simillarity.

curious_mind
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    @Adriano I believe he means that the points on the line are $B, A, X$ in that order, IE that $X$ lies on the extended line. – Calvin Lin Aug 29 '13 at 19:25

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Hint: Construct point $C'$ on $BA$ extended, such that $AC= AC'$.

Prove that $AP$ is the perpendicular bisector of $CC'$. Hence $PC=PC'$.

Hint: Apply triangle inequality to triangle $PBC'$.

Calvin Lin
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  • How you have got this idea?I mean it is verrrry hard for me. please tell me on what type of questions, we required constructions. THANK YOU. – curious_mind Aug 30 '13 at 03:36
  • @user91374 Think of how to relate the lines to each other. These kind of triangle inequality often require some reflection / rotation / translation argument. We know that we must transform $AB+AC$ into a straight line, which motivates the construction of $C'$. We then need $PC' = PC$, and so use the definition of $P$. – Calvin Lin Aug 30 '13 at 07:25
  • INTELLEGNT!!!!! – curious_mind Aug 30 '13 at 12:36