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ღ kekio ღ

17/08/2017 at 08:29
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For the triangle loss in A = 90 degrees

M point in BC. drawing-MO perpendicular to AB

MK drawing perpendicular to ac. On the beam of rays of OM and KM in turn retrieved 2 points D and E for OM = OD, KM = KE

C/M AD = AE



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    Kaya Renger Coordinator 17/08/2017 at 08:40

    A B C M E D O K

    We have : 

    \(AB\perp DM\) at O

    OD = OM

    => \(\Delta ADM\) is a isosceles triangle at A (1)

    Similar with \(\Delta AME\)

    =>  \(\Delta ADM\) is a isosceles triangle at A (2)

    From (1) and (2) 

    => AD = AE = AM
     

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