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I have the following problem: In a triangle ABC, the measure of the angle formed by the external angle bisectors of B and C is equal to twice the measure of the angle A. Find out the value of angle A.

Can anyone suggest how to approach this example?

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  • $\begingroup$ @FernadoGómez I assume you mean the external angle bisectors? $\endgroup$
    – Toby Mak
    Feb 17, 2019 at 2:30
  • $\begingroup$ @TobyMak yes, external angle bisectors, sorry, had to translate it from Spanish. I'll edit my post. $\endgroup$ Feb 17, 2019 at 2:35

1 Answer 1

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Hint:

Directly from the question, we know that:

$$\frac{180º - B}{2} = \frac{180º - C}{2} = 2A$$

and we also know that:

$$A+B+C = 180º$$

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  • $\begingroup$ Meaning then that: $$ \frac{180-B}{2}=2A B=180-4A $$ And: $$ \frac{180-C}{2}=2A C=180-4A $$ Hence: $$ A+180-4A+180-4A=180 \ A=\frac{180}{7} $$ Would that be the answer? $\endgroup$ Feb 17, 2019 at 2:37
  • $\begingroup$ @FernandoGómez Yes, that's correct. $\endgroup$
    – Toby Mak
    Feb 17, 2019 at 2:37
  • $\begingroup$ Thank you loads. $\endgroup$ Feb 17, 2019 at 2:39

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