Chapter 15, Problem 19RP

To determine

To Arrange: $\text{The sides of a triangle in order from longest to shortest}\text{.}$

Answer to Problem 19RP

  $\text{BD<AD<CD<BC=AB<AC}$

Explanation of Solution

Given:

  

Formula Used:

Angle -Sum Property: It states that in any triangle the sum of internal angles of a triangle is always ${180}^o$

In a triangle angle opposite to larger side is always larger, and smallest side is smallest.

  $\begin{array}{l}If\\ a>b\\ b>c\\ then\\ a>c\end{array}$

Calculation:

  $\begin{array}{l}In\Delta ACD\\ \angle ACD=180-\angle A-\angle ADC\\ \angle ACD=180-58-93\\ \angle ACD={29}^0\\ \angle BCD=\angle ACD={29}^0\\ \text{In }\Delta \text{BCD}\\ \angle B=180-\angle A-2\left(\angle ADC\right)\\ \angle B=180-58-2\left(29\right)\\ \angle B={64}^0\\ \angle CDB=180-\angle ADC\\ \angle CDB=180-93={87}^0\\ \text{Greater sides corresponding to greater angles}\text{.}\\ \angle ACD<\angle A<\angle ADC\\ \text{⇒}AD<CD<AC\\ \angle BCD<\angle B<\angle CDB\\ \text{⇒}BD<CD<BC\\ \angle A=\angle C<\angle B\\ \text{⇒}BC=AB<AC\\ \text{by using Angle bisector theorem}\\ \frac{AD}{BD}=\frac{AC}{BC}\\ \text{⇒}BC<AC\\ \text{⇒}BC<AD\\ \text{arranging the orders}\\ \text{BD<AD<CD<BC=AB<AC}\end{array}$

Conclusion:

Thus, the arrangement is $\text{BD<AD<CD<BC=AB<AC}$