Question

in trapezoid mode, diagonal \\( \overline{md} \\) is perpendicular to \\( \overline{om} \\), and \\( om = me = ed = 5 \\) in. find \\( m\angle odm, m\angle mod \\), and \\( m\angle med \\). answer: \\( m\angle odm = 45^{circ} \\) \\( m\angle mod = 45^{circ} \\) \\( m\angle med = 90^{circ} \\)

Explanation:

Step1: Analyze triangle \(OMD\)

Since \(OM = ED = 5\) in and \(MODE\) is a trapezoid (\(OM\parallel ED\)), and \(MD\perp OM\). In right - triangle \(OMD\), \(OM = ED\) and \(OM\parallel ED\) implies \(OM = MD\) (by the properties of the trapezoid and given equal side lengths). In a right - triangle \(OMD\) with \(OM = MD\), using the property that in a right - isosceles triangle \(\angle ODM=\angle MOD\).
We know that for a right - triangle \(A + B=90^{\circ}\) (where \(A\) and \(B\) are the non - right angles) and if \(A = B\) (because \(OM = MD\)), then \(\angle ODM=\angle MOD=\frac{90^{\circ}}{2}=45^{\circ}\).

Step2: Analyze \(\angle MED\)

Since \(OM\parallel ED\) and \(MD\perp OM\), by the property of parallel lines (if a line is perpendicular to one of two parallel lines, it is perpendicular to the other). So \(MD\perp ED\). Also, \(ME = ED = 5\) in. In quadrilateral \(MED\), \(\angle MED = 180^{\circ}-\angle EDM\). Since \(\angle EDM = 90^{\circ}\) (from \(MD\perp ED\)) and \(ME = ED\) (isosceles right - triangle property in the context of the trapezoid's side - length relationships), \(\angle MED=90^{\circ}\) (using the sum of angles in a quadrilateral and the properties of parallel and perpendicular lines in the trapezoid).

Answer:

\(m\angle ODM = 45^{\circ}\), \(m\angle MOD = 45^{\circ}\), \(m\angle MED=90^{\circ}\)