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Exterior angle bisectors of the side $\triangle ABC$ at vertices $B$ and $C$ intersect at $D$. Find $\angle BDC$ if $\angle BAC=40^{\circ}$

I cannot visualize this problem... If I draw a triangle and bisect the exterior angles, they never meet at a common point. Is this some sort of typo?

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  • $\begingroup$ Well, that or you're working in spherical geometry. $\endgroup$ – Dan Uznanski Sep 18 '14 at 4:07
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    $\begingroup$ The angle bisectors are probably to be thought of as lines through $B$ and $C$, not rays. $\endgroup$ – Blue Sep 18 '14 at 4:08
  • $\begingroup$ @Blue though in that case there would be no meaningful distinction between external and internal: both describe the same line. $\endgroup$ – Dan Uznanski Sep 18 '14 at 4:10
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    $\begingroup$ @DanUznanski: Internal angle bisector lines pass through the interior of the triangle; exterior angle bisector lines ---that is, lines bisecting the exterior angles--- do not. The interior bisector at a vertex is in fact perpendicular to the external bisector at that vertex. Definitely not the same line. $\endgroup$ – Blue Sep 18 '14 at 4:12
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    $\begingroup$ Oh, I see, it bisects the angle between the usual part of one edge and the extension of the other; for some reason I thought we were talking about both extensions. Silly me! $\endgroup$ – Dan Uznanski Sep 18 '14 at 4:20
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Hope the following sketch can help.

enter image description here

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  • $\begingroup$ I see so my picture was drawn with the rays pointing in the opposite direction. Wouldn't angle ACD be 90 if you're bi secting a line $\endgroup$ – adam Sep 18 '14 at 4:22
  • $\begingroup$ @adam It would be better if you consider external bisectors as lines instead of rays. $\endgroup$ – Mick Sep 18 '14 at 4:28
  • $\begingroup$ So the key to solving this problem is to know that interior angle bisectors are prepindicular to external angle bisectors. So I just draw a interior angle bisector and that will be perpendicular to line AC. Will this work? $\endgroup$ – adam Sep 18 '14 at 4:30
  • $\begingroup$ That will work. Another approach (it is simpler, I think) is, apply the combination of "angle sum of triangle" + "adjacent angles on a straight line". $\endgroup$ – Mick Sep 18 '14 at 4:35
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    $\begingroup$ Let's call $C=\angle ACB, B=\angle ABC$. Clearly $\angle CBD=90^o-B/2$ and $\angle BCD=90^o-C/2$ therefore $\angle CDB=(180^o-\angle CBD-\angle BCD)=(B+C)/2=(180-A)/2=140^o/2=70^o$ so I was not kidding for the one line answer. $\endgroup$ – user175968 Sep 18 '14 at 4:42
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Exterior angle bisectors

The exterior angle bisectors are just orthogonal to the interior angle bisectors, hence $$\widehat{BDC}=\pi-\widehat{BIC}=\frac{\widehat{ABC}+\widehat{ACB}}{2}=\frac{\pi-\widehat{BAC}}{2}.$$ This gives that if $\widehat{BAC}=40^\circ$, then $\widehat{BDC}=70^\circ$.

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It cause you draw the bisector of the small sideways. for example if b>c. you must continue the AC for external angel. This problem came out of this equation: Triangle Image

AB/AC = BE/EC

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