Chapter 15.3, Problem 25PSC

To determine

To conclude: that $\angle 1>\angle 4$ as per given condition.

Answer to Problem 25PSC

  $XV>VY$ .

Explanation of Solution

Given:

A triangle in which $\overline{AD}$ is the median and $\angle ABD>\angle ACD$

as shown in the given figure:

  

Calculation:

According to the question, a triangle in which $\overline{AD}$ is the median and $\angle ABD>\angle ACD$ . Now,

  $\begin{array}{l}\overline{BD}\cong \overline{BC}\left\{\text{median from the vertex bisects the opp}\text{. side}\right\}\\ \angle ABD>\angle ACD\left\{\text{given}\right\}\\ AC>AB\\ \overline{AD}\cong \overline{AD}\left\{\text{reflexive property}\right\}\\ \angle 3>\angle 2\left\{\text{converse of hinge theorem}\right\}\\ \overline{ED}\cong \overline{ED}\left\{\text{reflexive property}\right\}\\ EC>EB\left\{\text{converse of hinge theorem}\right\}\\ \overline{ZV}\cong \overline{ZV}\left\{\text{reflexive group}\right\}\\ \therefore \angle 1>\angle 4\end{array}$

Thus, $XV>VY$