Question

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 ce 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:

• For the first blank: Since \(CE\) is the perpendicular line segment from \(C\) to \(AB\), and \(BE\) is a part of \(AB\) (a line - segment), by the shortest - distance theorem (the perpendicular distance is the shortest distance from a point to a line), \(BE\) is the shortest distance from \(B\) to \(CE\).
• For the second blank: Similarly, \(AE\) is the shortest distance from \(A\) to \(CE\) because of the shortest - distance theorem (the perpendicular distance is the shortest distance from a point to a line).
• For the third blank: \(BE + AE=BA\) because of the segment addition postulate (if \(E\) lies on \(AB\), then \(AB = AE + EB\)).

Answer:

1. \(CE\)
2. \(AE\)
3. segment addition postulate