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The side BC of TRIANGLE ABC …

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The side BC of TRIANGLE ABC IS PRODUCED TO D the biscetor of angle a meets BC in L prove that angle ABC + Angle ACD=2angle ALC

    • Posted by Akshat Pandey 6 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Sia ? 6 years, 6 months ago

    {tex}\bigtriangleup{/tex}ABC,
    Ext.{tex}\angle{/tex}ACD = {tex}\angle{/tex}ABC + {tex}\angle{/tex}BAC (Exterior Angle Theorem)

    {tex}\Rightarrow{/tex} {tex}\angle{/tex}ABC + {tex}\angle{/tex}ACD = {tex}\angle{/tex}ABC + {tex}\angle{/tex}ABC + {tex}\angle{/tex}BAC
    (Adding {tex}\angle{/tex}ABC to both sides)
    = 2 {tex}\angle{/tex}ABC + {tex}\angle{/tex}BAC = 2{tex}\angle{/tex}ABC + 2{tex}\angle{/tex}BAL
    (AL is the bisector of {tex}\angle{/tex}BAC)
    = 2({tex}\angle{/tex}ABC + {tex}\angle{/tex}BAL) = 2{tex}\angle{/tex}ALC
    (because, In{tex}\bigtriangleup{/tex}ABL, Ext.{tex}\angle{/tex}ALC = {tex}\angle{/tex}ABC + {tex}\angle{/tex}BAL) (Exterior Angle Theorem)

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