Let P be any point inside a triangle $\triangle ABC$. $A$ and $P$ are joined and extended so that $AP$ when extended intersects $BC$ on $D$. Similarly define $E$ and $F$ on $CA$ and $AB$. Prove that $PD + PE + PF < \max(AB , BC , CA)$.
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We have $$\frac{PD}{AD}+\frac{PE}{BE}+\frac{PF}{CF}=1.$$ But $AD,BE,CF<\max \{AB,BC,AC\}.$ Then $$PD + PE + PF < \max(AB , BC , CA).$$
Roman83
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