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  • #62962
    Pinaki Biswas
    Participant

    In a triangle ABC, right-angled at A, the altitude through A and the internal bisector through A have lengths 3 and 4 respectively. Find the length of the median through A.

     

    #63344
    Saumik Karfa
    Participant

    Here $AI$ is the internal bisector of $A$ and $AL$ is the altitude through $A$.

    Now We know that if $AM$ is the median through $A$ then $AM=BM=MC=\frac12 BC$

    In $\triangle ALI$,

    $LI^2=4^2-3^2$

    $\Rightarrow LI=\sqrt 7$

    therefore $\tan \theta = \frac{\sqrt 7}{3}$

    From the figure we can see that $\angle LAB=\frac{\pi}{4}+\theta$ and $\angle LAC=\frac{\pi}{4}-\theta$

    Now from $\triangle ALB$ and $\triangle ALC$ the value of $BL$ and $LC$ can easily be obtained.

    and Median $=\frac12 BC = \frac 12 [BL+LC]$

    DONE!!!!

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