I为内心,P在AB边上,若IP交BC于Q,求QC

tommywong

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设I为三角形ABC内心，三边长AB=7,BC=6,AC=5,P在AB 边上,且AP=2,若IP交BC于Q,求QC

kuing

熟知 $a\vv{IA}+b\vv{IB}+c\vv{IC}=\vv0$，即 $a\bigl(\vv{BA}-\vv{BI}\bigr)-b\vv{BI}+c\bigl(\vv{BC}-\vv{BI}\bigr)=\vv0$，故
\[\vv{BI}=\frac{a\vv{BA}+c\vv{BC}}{a+b+c},\]
代入数据即
\[\vv{BI}=\frac13\vv{BA}+\frac7{18}\vv{BC},\]
设$QC=x$，则
\begin{align*}
\vv{BP}&=\frac57\vv{BA},\\
\vv{BQ}&=\frac{6-x}6\vv{BC},
\end{align*}
故由$P$, $I$, $Q$三点共线应有
\[\frac13\cdot\frac75+\frac7{18}\cdot\frac6{6-x}=1,\]
解得
\[x=\frac{13}8.\]

aipotuo

\[\dfrac{S_{\triangle{BPQ}}}{S_{\triangle{ABC}}}=\dfrac{BQ\cdot BP}{BA\cdot BC}=\dfrac{1/2r(BP+BQ)}{1/2r(AB+BC+CA)}\Longrightarrow
\dfrac{1}{BP}+\dfrac{1}{BQ}=\dfrac{3}{7}\].