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

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.

• Hint: find its intercept on $BC$ and one other point known to be on the bisector. – dxiv Sep 19 '16 at 21:59
• So, are you saying to position the triangle in a cartesian plane? – Lily Sep 19 '16 at 22:00
• How does one known angle bisector would help me find the angle bisector of vertex A? Should I look at the concurrency? – Lily Sep 19 '16 at 22:02
• 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. – dxiv Sep 19 '16 at 22:04
• 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? – Lily Sep 19 '16 at 22:10

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.

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:

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:

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 $K$.
• How did you get P and Q? – Lily Sep 19 '16 at 22:05
• @Lily: by intersecting two lines parallel to $AB$ and $AC$ at the same distance from $AB$ and $AC$. – Jack D'Aurizio Sep 19 '16 at 22:10
• 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. – Lily Sep 19 '16 at 22:16
• @Lily: that follows from the definition of angle bisector as a locus. – Jack D'Aurizio Sep 19 '16 at 22:21
• Ok. Thank you so much for your help. I learned a lot. – Lily Sep 19 '16 at 22:21