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O is the centreof the circle …

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O is the centreof the circle and BCD is a tangent to it at C. Prove that angle ABC + angle ACD= 90°

    • Posted by Naman Sen 6 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Sia ? 6 years, 6 months ago


    {tex}\angle OCD = 90^\circ{/tex} (tangent and radii are {tex}\bot {/tex} to one another at the point of contact)
    In {tex}\triangle{/tex}OCA,
    OC = OA (radii of circle)
    Hence, {tex}\angle OCA = \angle OAC{/tex} (angles opposite to equal sides are equal)
    Also, {tex}\angle OCD = \angle OCA + \angle ACD{/tex}
    {tex}90^\circ = \angle OAC + \angle ACD{/tex} {tex}\left( {\because \angle OCA = \angle OAC} \right){/tex}
    {tex}90^\circ = \angle BAC + \angle ACD{/tex}
    Hence, {tex}\angle BAC + \angle ACD = 90^\circ{/tex}
    Hence proved.

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