Giúp mình câu e đi ạ

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Giúp mình câu e đi ạ
giup-minh-cau-e-di-a

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Trung Dũng 4 years 2021-05-18T03:04:52+00:00 1 Answers 8 views 0

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  1. ThanhThu
    0
    2021-05-18T03:06:37+00:00
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    Giải thích các bước giải:

     a) Ta có:

    $\begin{array}{l}
    \left\{ \begin{array}{l}
    \widehat Bchung\\
    \widehat {AHB} = \widehat {CAB} = {90^0}
    \end{array} \right.\\
     \Rightarrow \Delta ABH \sim \Delta CBA\left( {g.g} \right)
    \end{array}$

    b) Ta có:

    $\begin{array}{l}
    \Delta ABH \sim \Delta CBA\left( {g.g} \right)\\
     \Rightarrow \dfrac{{AH}}{{CA}} = \dfrac{{AB}}{{CB}}\\
     \Rightarrow \dfrac{{AH}}{{AB}} = \dfrac{{AC}}{{BC}}
    \end{array}$

    c) Ta có:

    $\begin{array}{l}
    \left\{ \begin{array}{l}
    \widehat {AHB} = \widehat {CHA} = {90^0}\\
    \widehat {ABH} = \widehat {CAH}\left( { + \widehat {HAB} = {{90}^0}} \right)
    \end{array} \right.\\
     \Rightarrow \Delta AHB \sim \Delta CHA\left( {g.g} \right)\\
     \Rightarrow \dfrac{{AH}}{{CH}} = \dfrac{{HB}}{{HA}}\\
     \Rightarrow A{H^2} = CH.BH
    \end{array}$

    d) Ta có:

    $\begin{array}{l}
    \Delta ABC;\widehat A = {90^0};AB = 15cm;AC = 20cm\\
     \Rightarrow BC = \sqrt {A{B^2} + A{C^2}}  = 25cm
    \end{array}$

    Mà $\dfrac{{AH}}{{AB}} = \dfrac{{AC}}{{BC}} \Rightarrow AH = \dfrac{{AB.AC}}{{BC}} = 12cm$

    Lại có:

    $\begin{array}{l}
    \Delta ABH \sim \Delta CBA\left( {g.g} \right)\\
     \Rightarrow \dfrac{{BH}}{{BA}} = \dfrac{{AB}}{{BC}}\\
     \Rightarrow BH = \dfrac{{A{B^2}}}{{BC}} = 9cm\\
     \Rightarrow CH = BC – BH = 16cm\\
     \Rightarrow \dfrac{{BH}}{{CH}} = \dfrac{9}{{16}}
    \end{array}$

    Mà: $\dfrac{{DH}}{{AH}} = \dfrac{{6,75}}{{12}} = \dfrac{9}{{16}}$

    $ \Rightarrow \dfrac{{BH}}{{CH}} = \dfrac{{DH}}{{AH}}$

    Khi đó:

    $\begin{array}{l}
    \left\{ \begin{array}{l}
    \widehat {BHD} = \widehat {CHA} = {90^0}\\
    \dfrac{{BH}}{{CH}} = \dfrac{{DH}}{{AH}}
    \end{array} \right.\\
     \Rightarrow \Delta BHD \sim \Delta CHA\left( {g.g} \right)\\
     \Rightarrow \widehat {HBD} = \widehat {HCA}\\
     \Rightarrow BD//AC\\
     \Rightarrow BD \bot BA
    \end{array}$

    d) Ta có:

    $\begin{array}{l}
    \left\{ \begin{array}{l}
    \widehat {ABF} = \widehat {HBE} = \dfrac{1}{2}\widehat {ABC}\\
    \widehat {BAF} = \widehat {BHE} = {90^0}
    \end{array} \right.\\
     \Rightarrow \Delta ABF \sim \Delta HBE\left( {g.g} \right)\\
     \Rightarrow \widehat {AFB} = \widehat {HEB}\\
     \Rightarrow \widehat {AFE} = \widehat {AEF}\\
     \Rightarrow \Delta AEF \text{cân ở A}\\
     \Rightarrow AE = AF
    \end{array}$

    giup-minh-cau-e-di-a

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