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If $P$ is any point on a straight line drawn through the vertex $A$ of an isosceles triangle $ABC$, parallel to the base, prove that $PB+PC>AB+AC$

Please only give hint and tell how should I start

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  • $\begingroup$ What's the set of all points $Q$ such that $QB+QC = AB+AC$? $\endgroup$
    – Magma
    Feb 6, 2020 at 15:02
  • $\begingroup$ Alternatively you can give the points coordinates. Let $A = (x,y)$, $B = (-x,y)$, $C = (0,0)$ and $P = (a,0)$. Express both sides in terms of $x,y,a$, then a lot of rewriting the inequality will yield the desired result. $\endgroup$
    – Magma
    Feb 6, 2020 at 15:05

2 Answers 2

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Hint: reflect $B$ with respect to $AP$. Apply triangle inequality.

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Extend BA to C’ such that BA = AC’.

Prove that PC = PC’ by showing that AP is the perpendicular bisector of CC’.

Then, apply triangle inequality to $\triangle PBC’$.

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