Question

consider trapezoid lmno. what information would verify that lmno is an isosceles trapezoid? check all that apply. $overline{ln}congoverline{mo}$ $overline{lm}congoverline{on}$ $overline{lo}congoverline{mn}$ $angle lcongangle n$ $angle lcongangle m$

Explanation:

• For \(\overline{LO}\cong\overline{MN}\): In an isosceles trapezoid, the non - parallel sides (legs) are congruent. Here, \(LM\parallel ON\), so \(LO\) and \(MN\) are the non - parallel sides (legs). If \(LO\cong MN\), then by the definition of an isosceles trapezoid (a trapezoid with congruent non - parallel sides), \(LMNO\) is an isosceles trapezoid.
• For \(\overline{LN}\cong\overline{MO}\): In an isosceles trapezoid, the diagonals are congruent. If the diagonals \(LN\) and \(MO\) of trapezoid \(LMNO\) are congruent (\(LN\cong MO\)), then \(LMNO\) is an isosceles trapezoid.

• For \(\angle L\cong\angle M\): In a trapezoid \(LMNO\) with \(LM\parallel ON\), \(\angle L\) and \(\angle M\) are adjacent angles on the same base \(LM\). In a trapezoid (not necessarily isosceles), \(\angle L+\angle O = 180^{\circ}\) and \(\angle M+\angle N=180^{\circ}\) (since \(LM\parallel ON\)). If \(\angle L\cong\angle M\), then \(\angle O\cong\angle N\). Also, using the AAS (Angle - Angle - Side) congruence criterion for the triangles formed (e.g., if we consider triangles formed by the diagonals), we can show that the non - parallel sides are congruent.

• For \(\overline{LM}\cong\overline{ON}\): If \(LM\cong ON\) and \(LM\parallel ON\), then \(LMNO\) is a parallelogram (by the definition of a parallelogram: a quadrilateral with both pairs of opposite sides parallel and congruent), not an isosceles trapezoid.

• For \(\angle L\cong\angle N\): In a trapezoid \(LMNO\) with \(LM\parallel ON\), \(\angle L\) and \(\angle N\) are not base angles in the sense required for an isosceles trapezoid. In an isosceles trapezoid with \(LM\parallel ON\), the base angles are \(\angle L\) and \(\angle M\) (with respect to base \(LM\)) or \(\angle O\) and \(\angle N\) (with respect to base \(ON\)).

Answer:

\(\overline{LN}\cong\overline{MO}\), \(\overline{LO}\cong\overline{MN}\), \(\angle L\cong\angle M\)