Prove using analytic methods:

1. the measure of the line segment joining 2 sides of a triangle is equal to one-half to measure of the third side

2. the medians to the congruent sides of an isosceles triangle are congruent

Guest Nov 26, 2017

#1**+1 **

1) I am fairly certain that you are referring to the midsegment theorem. Here is a diagram for you to reference as I prove:

We know that by the midsegment theorem that the segment bisects both of its intersecting sides and is parallel to the third side. \(\angle CDE\cong\angle CAB\) and \(\angle CED\cong\angle CBA\) by the Corresponding Angles Postulate. By the Angle-Angle Similarity Theorem, \(\triangle CDE \sim \triangle CAB\). We know that \(\frac{CD}{CA}\) equals some ratio. We know that \(\overline{CA}\) is broken up into two smaller segments, \(\overline{CD}\) and \(\overline{DA}\). The ratio is now \(\frac{CD}{CD+DA}\). By the given info, \(CD=DA\), so, by the substitution property of equality, the ratio can be written as\(\frac{CD}{CD+CD}=\frac{CD}{2CD}=\frac{1}{2}\). Since the ratio of corresponding sides of a similar triangle are equal, the ratio of \(\frac{DE}{AB}=\frac{1}{2}\). Rewriting this ratio turns into \(2DE=AB\Rightarrow DE=\frac{1}{2}AB\).

2) Here is a diagram again!

In this diagram here, \(AB=AC\), and \(D\) is the midpoint of \(\overline{AB}\) and \(E\) is the midpoint of \(\overline{AC}\), which makes \(\overline{BE}\) and \(\overline{CD}\) medians of this isosceles triangle.

By the given info, \(AB=AC\) and \(AD=AE\) because they are midpoints of equal-length segments. \(\angle A\cong\angle A\) by the reflexive property of congruence. Therefore, by Side-Angle-Side Congruency Theorem, \(\triangle ABE\cong\triangle ACD\). Utilizing the fact that corresponding parts of congruent triangles are congruent, \(CD=BE\). We are done!

TheXSquaredFactor
Nov 26, 2017