Sửa đề câu d: $BH.BQ + CH.CK = B{C^2}$

Giải thích các bước giải:

 a) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat Achung\\
\widehat {AQB} = \widehat {AKC} = {90^0}
\end{array} \right.\\
 \Rightarrow \Delta ABQ \sim \Delta ACK\left( {g.g} \right)
\end{array}$

b) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {HKB} = \widehat {HQC} = {90^0}\\
\widehat {KHB} = \widehat {QHC}\left( {dd} \right)
\end{array} \right.\\
 \Rightarrow \Delta KHB \sim \Delta QHC\left( {g.g} \right)\\
 \Rightarrow \dfrac{{HK}}{{HQ}} = \dfrac{{HB}}{{HC}}\\
 \Rightarrow HK.HC = HB.HQ
\end{array}$

c) Ta có:

$\begin{array}{l}
\Delta KHB \sim \Delta QHC\left( {g.g} \right)\\
 \Rightarrow \dfrac{{HK}}{{HQ}} = \dfrac{{HB}}{{HC}}\\
 \Rightarrow \dfrac{{HK}}{{HB}} = \dfrac{{HQ}}{{HC}}
\end{array}$

Khi đó:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {KHQ} = \widehat {BHC}\left( {dd} \right)\\
\dfrac{{HK}}{{HB}} = \dfrac{{HQ}}{{HC}}
\end{array} \right.\\
 \Rightarrow \Delta HQK \sim \Delta HBC\left( {c.g.c} \right)
\end{array}$

d) Kẻ đường cao $AD$ ($D\in BC$)

Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat Bchung\\
\widehat {BDH} = \widehat {BQC} = {90^0}
\end{array} \right.\\
 \Rightarrow \Delta BDH \sim \Delta BQC\left( {g.g} \right)\\
 \Rightarrow \dfrac{{BH}}{{BC}} = \dfrac{{BD}}{{BQ}}\\
 \Rightarrow BH.BQ = BC.BD\left( 1 \right)
\end{array}$

Lại có

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat Cchung\\
\widehat {CDH} = \widehat {CKB} = {90^0}
\end{array} \right.\\
 \Rightarrow \Delta CDH \sim \Delta CKB\left( {g.g} \right)\\
 \Rightarrow \dfrac{{CD}}{{CK}} = \dfrac{{CH}}{{CB}}\\
 \Rightarrow CH.CK = CD.CB\left( 2 \right)
\end{array}$

Từ $\left( 1 \right),\left( 2 \right) \Rightarrow BH.BQ + CH.CK = BC.BD + CD.CB = B{C^2}$