Kaya Renger Coordinator
05/11/2017 at 11:01-
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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