Question

in the diagram, \\( \angle gef \\) and \\( \angle hfd \\) are exterior angles of \\( \triangle def \\); \\( m\angle gef = 67.5 ^ { \circ } \\); and \\( m\angle hfd = 146.25 ^ { \circ } \\).
prove that \\( \triangle def \\) is isosceles by choosing from the options provided in each row.
\\( \

$$\begin{array} { | c | } \\hline \\text { statement } \\\\ \\hline \\begin{array} { c } { \\triangle d e f ; \\text { exterior angles } \\angle g e f \\text { and } \\angle h f d } \\\\ { ; m \\angle g e f = 67.5 ^ { \\circ } ; \\text { and } } \\\\ { m \\angle h f d = 146.25 ^ { \\circ } } \\end{array}$$

\\\\ \hline \

$$\begin{array} { c } { \\angle d e f \\text { and } \\angle g e f \\text { are supplementary } } \\\\ { \\text { angles } } \\end{array}$$

\\\\ \hline m \angle d e f + m \angle g e f = 180 ^ { \circ } \\\\ \hline m \angle d e f + 67.5 ^ { \circ } = 180 ^ { \circ } \\\\ \hline m \angle d e f = 112.5 ^ { \circ } \\\\ \hline m \angle e d f + m \angle d e f = m \angle h f d \\\\ \hline m \angle e d f + 112.5 ^ { \circ } = 146.25 ^ { \circ } \\\\ \hline m \angle e d f = 33.75 ^ { \circ } \\\\ \hline m \angle e d f + m \angle d e f + m \angle d f e = 180 ^ { \circ } \\\\ \hline 33.75 ^ { \circ } + 112.5 ^ { \circ } + m \angle d f e = 180 ^ { \circ } \\\\ \hline m \angle d f e = 33.75 ^ { \circ } \\\\ \hline \overline { d e } \cong \overline { f e } \\\\ \hline \triangle d e f \text { is isosceles } \\\\ \hline \end{array} \\)

Explanation:

Step1: Find \(m\angle DEF\)

Since \(\angle DEF\) and \(\angle GEF\) are supplementary angles, \(m\angle DEF + m\angle GEF=180^{\circ}\). Substituting \(m\angle GEF = 67.5^{\circ}\), we get \(m\angle DEF+67.5^{\circ}=180^{\circ}\). Then \(m\angle DEF=180^{\circ}- 67.5^{\circ}=112.5^{\circ}\).

Step2: Find \(m\angle EDF\)

Using the exterior - angle theorem (\(m\angle EDF + m\angle DEF=m\angle HFD\)), substituting \(m\angle DEF = 112.5^{\circ}\) and \(m\angle HFD = 146.25^{\circ}\), we have \(m\angle EDF+112.5^{\circ}=146.25^{\circ}\). So \(m\angle EDF=146.25^{\circ}-112.5^{\circ}=33.75^{\circ}\).

Step3: Find \(m\angle DFE\)

Using the angle - sum property of a triangle (\(m\angle EDF + m\angle DEF+m\angle DFE = 180^{\circ}\)), substituting \(m\angle EDF = 33.75^{\circ}\) and \(m\angle DEF = 112.5^{\circ}\), we get \(33.75^{\circ}+112.5^{\circ}+m\angle DFE=180^{\circ}\). Then \(m\angle DFE=180^{\circ}-(33.75^{\circ}+112.5^{\circ})=33.75^{\circ}\).

Step4: Prove \(\triangle DEF\) is isosceles

Since \(m\angle EDF=m\angle DFE = 33.75^{\circ}\), by the converse of the base - angles theorem (if two angles of a triangle are congruent, then the sides opposite those angles are congruent), \(\overline{DE}\cong\overline{FE}\). So \(\triangle DEF\) is isosceles.

Answer:

\(\triangle DEF\) is isosceles.