Chapter 15.3, Problem 3PSA

To determine

To Find: The longer side BC or DC for the given quadrilateral.

Answer to Problem 3PSA

  $\overline{DC}$

Explanation of Solution

Given:

  $\begin{array}{l}\to \overline{AB}\parallel \overline{CD}\\ \to \overline{AB}\cong \overline{AD}\\ \to \angle DAC={75}^o\\ \to \angle DCA={45}^o\end{array}$

  

Concept:

The Hinge Theorem: It states that if two sides of two triangles are congruent and angle of one is greater than the angle of second then the third side of the triangle will be greater than the another.

Calculation:

  $\begin{array}{l}\to \overline{AB}\parallel \overline{CD}\\ \to \overline{AB}\cong \overline{AD}\\ \to \angle DAC={75}^o\\ \to \angle DCA={45}^o\\ \text{⇒}m\angle ADC=180-75-45=60\left(\Delta \text{ Angle-sum theorem}\right)\\ \text{⇒}m\angle ACB=75\text{ and }m\angle BAC=45\left(\text{Alternate interior angle}\right)\\ \text{⇒}m\angle ABC=180-75-45=60\left(\Delta \text{ Angle-sum theorem}\right)\\ \text{⇒}\overline{DC}\text{ is longer }\left(\text{Opposite side of larger }\angle \text{ is larger}\right)\end{array}$

Conclusion:

Thus the $\overline{DC}$ is longer