Cho `\Delta ABC` nhọn ngoại tiếp `(I)`, `MN ⊥ AI=I` `(M\in AB, N\in AC)`. CMR: a) `((IB)/(IC))^2=(BM)/(CN)` B) `(BM)/(AB)+(CN)/(AC)+(AI^2)/(AB. AC)=1`

a)

Có: $\widehat{BIM}=\widehat{BIA}-\widehat{MIA}$

$\Rightarrow \widehat{BIM}=\left( 180{}^\circ -\widehat{IAB}-\widehat{IBA} \right)-90{}^\circ $

$\Rightarrow \widehat{BIM}=90{}^\circ -\dfrac{1}{2}\left( \widehat{A}+\widehat{B} \right)$

$\Rightarrow \widehat{BIM}=\dfrac{180{}^\circ -\left( \widehat{A}+\widehat{B} \right)}{2}$

$\Rightarrow \widehat{BIM}=\dfrac{\widehat{C}}{2}$

$\Rightarrow \widehat{BIM}=\widehat{ICN}=\widehat{BCI}$

Chứng minh tương tự:

Có: $\widehat{CIN}=\widehat{IBM}=\widehat{CBI}$

$\Rightarrow \Delta BIM\backsim\Delta ICN\backsim\Delta BCI\left( g.g \right)$

Từ: $\Delta BIM\backsim\Delta ICN\Rightarrow \dfrac{IB}{IC}=\dfrac{BM}{IN}$

Từ: $\Delta BCI\backsim\Delta ICN\Rightarrow \dfrac{IB}{IC}=\dfrac{IN}{CN}$

$\Rightarrow \dfrac{IB}{IC}\cdot \dfrac{IB}{IC}=\dfrac{BM}{IN}\cdot \dfrac{IN}{CN}$

$\Rightarrow {{\left( \dfrac{IB}{IC} \right)}^{2}}=\dfrac{BM}{CN}$

b)

$\Delta AMN$ có $AI$ vừa là đường cao, đường phân giác

$\Rightarrow \Delta AMN$ cân tại $A$

$\Rightarrow AM=AN$   và   $IM=IN$

Từ $\Delta BIM\backsim\Delta ICN$

$\Rightarrow IM.IN=BM.CN$

$\Rightarrow I{{M}^{2}}=BM.CN$

Biển đổi tương đương từ yêu cầu đề bài:

Từ:  $\dfrac{BM}{AB}+\dfrac{CN}{AC}+\dfrac{A{{I}^{2}}}{AB.AC}=1$

$\Leftrightarrow BM.AC+CN.AB+A{{M}^{2}}-I{{M}^{2}}=AB.AC$

$\Leftrightarrow BM.AC+CN.AB+A{{M}^{2}}-BM.CN=AB.AC$

$\Leftrightarrow A{{M}^{2}}+\left( CN.AB-BM.CN \right)=\left( AB.AC-BM.AC \right)$

$\Leftrightarrow A{{M}^{2}}+CN\left( AB-BM \right)=AC\left( AB-BM \right)$

$\Leftrightarrow A{{M}^{2}}+CN.AM=AC.AM$

$\Leftrightarrow A{{M}^{2}}=AM\left( AC-CN \right)$

$\Leftrightarrow A{{M}^{2}}=AM.AN$ (luôn đúng do $AM=AN$)

Vậy: $\dfrac{BM}{AB}+\dfrac{CN}{AC}+\dfrac{A{{I}^{2}}}{AB.AC}=1$