Question

geometry with data analysis ic qtr b young continuous learning center - credit bearing (tutor)
triangles and their side lengths
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:

1. The first blank: Since we are drawing a perpendicular from \(C\) to \(AB\) (intersection \(E\)), and we know that the perpendicular distance is the shortest distance. For the distance from \(B\) to \(CE\), \(BE\) is the perpendicular segment (so \(CE\) is the line).
2. The second blank: For the distance from \(A\) to \(CE\), \(AE\) is the perpendicular segment (so \(AE\) is the shortest distance from \(A\) to \(CE\) by the shortest - distance theorem).
3. The third blank: \(BE + AE=BA\) because of the segment - addition postulate (which states that if \(E\) is a point on segment \(AB\), then \(AE + EB=AB\)).

Answer:

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