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 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:

Step1: Fill first blank

The first blank refers to the object that BE is the shortest distance to. Since BE is a segment from B to CE (the perpendicular), the blank is "CE".

Step2: Fill second blank

The second blank is the shortest distance from A to CE. By the shortest distance theorem, this is AE. So the blank is "AE".

Step3: Fill third blank

BE + AE equals BA because E lies on BA, so their sum is the whole segment. This uses the segment addition postulate. So the blank is "segment addition postulate".

Answer:

CE, AE, segment addition postulate