Chapter 9, Problem 29CUR

To determine

To prove: that the segment BE is greater than segment CD .

Explanation of Solution

Given:

Following figure is given

  

Also it is given that side AB is greater than side AC and segment BD is congruent to segment EC .

In the given figure, it is given that in $\Delta ABC$ side AB is greater than side AC , thus $\angle ACB$ will be greater than $\angle ABC$ .

Now consider the smaller triangles BDC and CEB .

  

Here, as it is mentioned in the question that $\overline{BD}\cong \overline{EC}$ and as side BC is common, the relation between the other side of two triangles depend on the angle opposite to those sides in their respective triangles respectively.

This is derived from the fact, that in a triangle if two side lengths are known, then the third side length depend on the angle inscribed between former two sides.

That is, in any two triangle which has 2 corresponding sides as congruent, than the relation between the other sides of triangles will depend on angle opposite to them. If the angles are equal then sides will be equal, if one angle is larger than other, than side of that triangle will be larger than corresponding side of other triangle.

Thus, in $\Delta BDC$ and $\Delta CEB$ , as the $\angle ECB$ is larger than $\angle DBC$ , thus the side opposite to them BE will be greater than side ED.

Thus, $BE>CD$ .

Conclusion:

Therefore, segment $BE>CD$ can be proved by the use of basic relation between side and angles of a triangle.