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: Recall the shortest distance theorem

The shortest distance from a point to a line is the perpendicular segment. So, \( BE \) is the shortest distance from \( B \) to \( CE \) (since \( CE \) is the perpendicular from \( C \) to \( AB \)), and \( AE \) is the shortest distance from \( A \) to \( CE \). This is by the "shortest distance from a point to a line is the perpendicular segment" theorem.

Step2: Apply inequality addition

We know \( BC > BE \) and \( AC > AE \) (from the shortest distance property). By the property of inequalities, if \( a > b \) and \( c > d \), then \( a + c > b + d \). So, \( BC + AC > BE + AE \).

Step3: Substitute \( BE + AE \) with \( BA \)

Since \( E \) lies on \( AB \), \( BE + AE = BA \) (segment addition postulate: if a point \( E \) is on segment \( AB \), then \( AB = AE + EB \)).

Step4: Conclude the inequality

Substituting \( BE + AE \) with \( BA \) in \( BC + AC > BE + AE \), we get \( BC + AC > BA \).

Answer:

The proof shows \( BC + AC > BA \) using the shortest distance theorem (perpendicular segment is shortest) and inequality addition, followed by segment addition postulate for substitution.