Chapter 4.6, Problem 25PPS

(a)

To determine

To find: The measurement of $\angle ABC$ .

Answer to Problem 25PPS

The required angle $\angle ABC=65°$

Explanation of Solution

Given information: It is given that $\angle BAC=50°$ and $\overline{AB}\cong \overline{AC}$

  

Calculation: Since $\overline{AB}\cong \overline{AC}$ ,

Hence, $AB=AC$

  $\Delta ABC$ is an isosceles triangle.

  $\begin{array}{l}\angle ABC\cong \angle ACB\\ \angle ABC=\angle ACB\end{array}$

By using Triangle-Sum Theorem,

  $\begin{array}{l}\angle ABC+\angle ACB+\angle BAC=180°\\ 2\angle ABC+\angle BAC=180°\\ 2\angle ABC=180°-50°\left[\because \angle BAC=50°\right]\\ \angle ABC=\frac{130°}{2}\\ \angle ABC=65°\end{array}$

(b)

To determine

To prove: The triangle $AED$ is an isosceles.

Explanation of Solution

Given information: It is given that $\overline{BE}\cong \overline{CD}$

  

Proof: Since $\overline{BE}\cong \overline{CD}$ ,

Hence, $BE=CD$

From part (a), $AB=AC$

Then,

  $\begin{array}{l}AE=AB+BE\\ AD=AC+CD\\ \therefore AE=AD\end{array}$

Therefore, the triangle $AED$ is an isosceles triangle.

Hence, proved.

(c)

To determine

To prove: The triangle $AED$ is equilateral.

Explanation of Solution

Given information: It is given that $\overline{BC}\parallel \overline{ED}\&\overline{ED}\cong \overline{AD}$

  

Proof: Since $\overline{BC}\parallel \overline{ED}$ ,

Hence,

  $\begin{array}{l}\angle ABC\cong \angle AED\\ \angle ACB\cong ADE\\ \therefore \angle AED\cong \angle ADE\end{array}$

Given $\overline{ED}\cong \overline{AD}$

  $\therefore ED=AD$

Thus, $AE=AD=DE$

Therefore, the triangle $AED$ is an equilateral triangle.

(d)

To determine

To find: The minimum information that is needed to prove $\Delta ABC\cong \Delta JKL$ .

Answer to Problem 25PPS

The minimum information that is needed is $\angle BAC\cong \angle KJL\text{and}\overline{AB}\cong \overline{JL}$ .

Explanation of Solution

Given information: It is given that $\Delta JKL$ is isosceles.

  

Proof: Since $\Delta ABC\cong \Delta JKL$ ,

Since triangle are isosceles, hence by side angle side congruence postulate,

  $\begin{array}{l}\overline{AB}\cong \overline{JL}\\ \angle BAC\cong \angle KJL\\ \overline{AC}\cong \overline{JK}\end{array}$