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Assume that a part of ∆ABC around vertex A is not visible. Describe how to find the angle bisector of ∠CAB.

I have no idea how to begin this problem. Any help is appreciated. Thank you.

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  • $\begingroup$ Hint: find its intercept on $BC$ and one other point known to be on the bisector. $\endgroup$ – dxiv Sep 19 '16 at 21:59
  • $\begingroup$ So, are you saying to position the triangle in a cartesian plane? $\endgroup$ – Lily Sep 19 '16 at 22:00
  • $\begingroup$ How does one known angle bisector would help me find the angle bisector of vertex A? Should I look at the concurrency? $\endgroup$ – Lily Sep 19 '16 at 22:02
  • $\begingroup$ Cartesian plane - not necessarily, what I say is that you can find those two points geometrically without using the hidden vertex $A$. Concurrency - yes, that's relevant. $\endgroup$ – dxiv Sep 19 '16 at 22:04
  • $\begingroup$ So you are saying to find the angle bisectos of angle B and angle C. Then their intersection will indicate the angle bisector of angle A? $\endgroup$ – Lily Sep 19 '16 at 22:10
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You may exploit the following property of the angle bisector through $A$: for any point $P$ on it, the distance of $P$ from the $AB$-side equals the distance of $P$ from the $AC$-side. Just find two distinct points with such property in the visible part and join them to get the wanted line.

enter image description here

You may take $P=I$ as the intersection of the internal angle bisectors from $B$ and $C$ and $Q=I_A$ as the intersection of the external angle bisectors from $B$ and $C$, for instance.

An alternative approach through triangle similarities: enter image description here

  1. Take $B'\in AB$ close to $B$ and let $C'\in AC$ such that $B'C'\parallel BC$;
  2. Let $C''\in BC$ be such that $B' C''\parallel AC$;
  3. Let $J$ the point in which the angle bisector of $\widehat{BB'C''}$ meets the $BC$-side
    and $K$ the projection of $J$ on $B'C''$;
  4. Let $L=BK\cap AC$ and $N\in BC$ be such that $LN\parallel KJ$;
  5. The parallel to $B'J$ through $L$ is the wanted angle bisector.

A third approach:

enter image description here

  1. Let $B'$ and $C'$ like before, let $M$ be the midpoint of $B'C'$;
  2. Let $A'$ be the intersection between the parallel to $AC$ through $B'$
    and the parallel to $AB$ through $C'$;
  3. Let $J\in B'C'$ be on the angle bisector of $\widehat{B' A' C'}$;
  4. Let $K$ be the symmetric of $J$ with respect to $M$;
  5. The wanted line is the parallel to $A'J$ through $K$.
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  • $\begingroup$ How did you get P and Q? $\endgroup$ – Lily Sep 19 '16 at 22:05
  • $\begingroup$ @Lily: by intersecting two lines parallel to $AB$ and $AC$ at the same distance from $AB$ and $AC$. $\endgroup$ – Jack D'Aurizio Sep 19 '16 at 22:10
  • $\begingroup$ or any point P on it, the distance of P from the AB-side equals the distance of P from the AC-side. How do we know that? This might be a stupid question. $\endgroup$ – Lily Sep 19 '16 at 22:16
  • $\begingroup$ @Lily: that follows from the definition of angle bisector as a locus. $\endgroup$ – Jack D'Aurizio Sep 19 '16 at 22:21
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    $\begingroup$ Ok. Thank you so much for your help. I learned a lot. $\endgroup$ – Lily Sep 19 '16 at 22:21

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