We have to describe and sketch a line m in a plane ABC such that points A, B, and C are noncollinear and equidistant from the line. To do so, let's start by drawing three noncollinear points A, B, and C, and the triangle they form △ABC.
Now draw a line m passing through two of the midpoints of △ABC.
Note that, by definition of midpoint, AM_1=CM_1 and BM_2=CM_2.
m is equidistant from A and C
Let's draw the distance from A to m and from C to m. In the diagram below, ∠OM_1A and ∠PM_1C are vertical angles and thus they are congruent.
We can see above that two angles and a nonincluded side of △AOM_1 are congruent to two angles and the corresponding nonincluded side of △CPM_1. Hence △AOM_1 ≅ △CPM_1 by AAS and thus their corresponding parts are congruent. Therefore, AO=CP, meaning that m is equidistant from A and C.
m is equidistant from B and C
Let's draw the distance from B to m and from C to m. In the diagram below, ∠RM_2C and ∠QM_2B are vertical angles and thus they are congruent.
We can see above that two angles and a nonincluded side of △CRM_2 are congruent to two angles and the corresponding nonincluded side of △BQM_2. Hence △CRM_2 ≅ △BQM_2 by AAS and thus their corresponding parts are congruent. Therefore, BQ=CR, meaning that m is equidistant from B and C.
m is equidistant from A, B, and C
We have already seen that m is equidistant from A and C. We have also seen that m is equidistant from C and B. Hence, by transitive property, m is also equidistant from A and B.
d(m,A)=d(m,C) d(m,C)=d(m,B) ⇒ d(m,A)=d(m,B)
Therefore, m is equidistant from A, B, and C.
