Question

proving the triangle inequality theorem
given: abc is a triangle
prove: bc + ac > ba
in triangle abc, we can draw a perpendicular line segment from vertex c to segment ab. the intersection of ab and the perpendicular is called e.
we know that be is the shortest distance from b to and that is the shortest distance from a to cb because of the shortest distance theorem. therefore, bc > be and ac > ae. next, add the inequalities: bc + ac > be + ae. then, be + ae = ba because of the therefore, bc + ac > ba by substitution.

Explanation:

We know that \(BE\) is the shortest distance from \(B\) to \(CE\) (because the perpendicular segment from a point to a line is the shortest distance). And \(AE\) is the shortest distance from \(A\) to \(CE\) (by the shortest - distance theorem). Also, \(BE + AE=BA\) because of the segment - addition postulate (if \(E\) lies on \(AB\), then \(AB = AE + EB\)).

Answer:

First blank: \(CE\); Second blank: \(AE\); Third blank: segment - addition postulate.