Question

prove: \\( \triangle d e f \cong \triangle d g f \\)

Explanation:

Step1: Calculate the length of \( \overline{EF} \) and \( \overline{GF} \)

The coordinates of \( E(-3,0) \) and \( F(0,0) \). Using the distance formula \( d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \), for \( E(-3,0) \) and \( F(0,0) \), \( d_{EF}=\sqrt{(0+3)^2+(0 - 0)^2}=3 \).
The coordinates of \( G(3,0) \) and \( F(0,0) \). For \( G(3,0) \) and \( F(0,0) \), \( d_{GF}=\sqrt{(0 - 3)^2+(0 - 0)^2}=3 \).

Step2: Analyze the property of \( \overline{DF}\)

For any segment \( \overline{DF} \), \( \overline{DF}\cong\overline{DF} \) by the reflexive property of congruence.

Step3: Apply the Leg - Leg Congruence Theorem

Since \( \triangle DEF \) and \( \triangle DGF \) are right - triangles (legs on \( x \) and \( y\) - axis which are perpendicular), \( \overline{EF}\cong\overline{GF} \) (both have length 3) and \( \overline{DF}\cong\overline{DF} \) (reflexive property), then \( \triangle DEF\cong\triangle DGF \) by the Leg - Leg Congruence Theorem.

Answer:

The length of \( \overline{EF} \) is \( 3 \) units, the length of \( \overline{GF} \) is \( 3 \) units, \( \overline{DF}\cong\overline{DF} \) by the reflexive property, and \( \triangle DEF\cong\triangle DGF \).