Question

given: parallelogram dfgj, dj > df
prove: m∠gjf > m∠djf

1. click here to insert
2. m∠gjf > m∠djf

Explanation:

Step1: Use properties of parallelograms

In parallelogram \(DFGJ\), \(DF\parallel GJ\). So, \(\angle DJF=\angle FJG_1\) (alternate - interior angles).

Step2: Apply the triangle - angle relationship

In \(\triangle DJF\), since \(DJ > DF\), by the converse of the triangle - angle relationship (the larger side is opposite the larger angle), \(\angle DGF>\angle DJF\) (where \(\angle DGF\) is the angle opposite \(DJ\) and \(\angle DJF\) is the angle opposite \(DF\) in \(\triangle DJF\)). And \(\angle GJF=\angle DGF+\angle FJG_2\) (angle - addition postulate).

Answer:

1. \(DJ > DF\) (Given), in \(\triangle DJF\), \(\angle DGF>\angle DJF\) (converse of the triangle - angle relationship: larger side opposite larger angle), \(DF\parallel GJ\) (property of parallelogram), \(\angle DJF = \angle FJG_1\) (alternate - interior angles), \(\angle GJF=\angle DGF+\angle FJG_2\) (angle - addition postulate).
2. Substitute \(\angle FJG_2=\angle DJF\) into \(\angle GJF=\angle DGF+\angle FJG_2\), we get \(\angle GJF>\angle DJF\) (if \(a = b + c\) and \(c>0\), then \(a>b\))