Common Segment Theorem Part 2

The last post I made last year was the Common Segment Theorem. In this post, I am going to discuss the converse of this theorem. The proof of these two theorems are very similar.

common segment


Consider the figure above. In the previous post, the Common Segment Theorem states that

If \overline{AC} \cong \overline {BC} then, \overline{AB} \cong \overline {CD}. The following is its converse and the proof.


If two collinear segments adjacent to a common segment is congruent, then, the overlapping segments formed are congruent.


Based on the figure above, the theorem states that if \overline{AB} \cong \overline {CD}, then \overline{AC} \cong \overline{BD}.

\overline{AB} \cong \overline{CD} is given.

AB = CD by the Definition of Congruence. That is, if two segments are congruent, then their lengths are equal.

AC = AB + BC (1) by the Segment Addition Postulate. If B is between A and C, then AC = AB + AC.

BD = BC + CD,(2) again, by the Segment Addition Postulate.

AC =  CD + BC , substituted CD to AB in (2) *

Notice that both the right hand side of the equation in (1) and * are both BC + CD.

Therefore, by substitution and transitive property of equality, AC = BD.

So, by definition of congruence,

\overline{AC} \cong \overline{BD}. This is what we want to prove.


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