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Kaya Renger Coordinator

05/11/2017 at 11:01
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An triangle has height and median divide the angle at the top into 3 equal angles. Calculate all angle of that triangle .




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    Dao Trong Luan Coordinator 06/11/2017 at 07:41

    A B C M H P

    Paint \(MP\perp AC\) at P.

    Consider two right-triangles: \(\Delta HAM-and-\Delta PAM\), we have:

    - AM is the common edge \(\left(\text{hypothesis}\right)\)

    - \(\widehat{HAM}=\widehat{MAP}-\left(\text{hypothesis}\right)\)

    => \(\Delta HAM=\Delta PAM-\left(\text{hypotenuse - acute angle}\right)\)

    => HM = PM

    Consider two right-triangles: \(\Delta ABH-and-\Delta AMH\), we have:

    - AH is the common edge \(\left(\text{hypothesis}\right)\)

    - \(\widehat{BAH}=\widehat{MAH}\left(\text{hypothesis}\right)\)

    => \(\Delta ABH=\Delta AMH\left(\text{hypotenuse - acute angle}\right)\)

    => BH = MH

    \(\Rightarrow BM=MH=\dfrac{1}{2}BC=\dfrac{1}{2}CM\)

    \(\Delta MPC\) right have: \(MI=\dfrac{1}{2}CM\)

    \(\Rightarrow\widehat{C}=30^o\Rightarrow\widehat{HAC}=60^o\Rightarrow\widehat{BAC}=\dfrac{3}{2}\widehat{HAC}=90^o\)

    So \(\left[{}\begin{matrix}\widehat{A}=90^o\\\widehat{B}=60^o\\\widehat{C}=30^o\end{matrix}\right.\)

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