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# prove using analytic methods

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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
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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

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