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Thread: Geometry - triangle problem

  1. #1
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    Geometry - triangle problem

    ABC is a triangle in which $\displaystyle \angle B = 2 \angle C.$ If D is a point on BC such that AD bisects $\displaystyle \angle BAC and AB =CD$Prove that $\displaystyle \angle BAC = 72^{\circ}$

    Can we go this way :

    Let $\displaystyle \angle A = 2t ; \angle B = 2x ; \angle C = x$[/latex] ( as $\displaystyle \angle B = 2\angle C$ Let $\displaystyle \angle ADC = m ; \angle ADB =n$ such that $\displaystyle \angle m + \angle n = 180^{\circ}$

    Now in $\displaystyle \triangle ADB ; 2x+n+t =180^{\circ}$( since $\displaystyle \angle A = 2t$ and D is the bisector of $\displaystyle \angle BAC$ also $\displaystyle \angle m + \angle x + \angle DAC = 180^{\circ}$

    $\displaystyle \angle m = \angle n + \angle 2x$( as m is external angle which is equal to sum of the opposite interior angles)
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  2. #2
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    Re: Geometry - triangle problem

    Geometry - triangle problem-triangle-15-mar-13.png
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  3. #3
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    Re: Geometry - triangle problem

    thanks a lot...
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