If two parallel lines are intersected by a transversal, then pair of interior angle are

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To prove: GHML is a rectangle.Proof:Since \[AB\parallel CD\]\[\angle AGH = \angle DHG{\text{ }}\left( {{\text{Alternate interior angles}}} \right)\]Dividing both side by 2, we have\[ \Rightarrow \dfrac{1}{2}\angle AGH = \dfrac{1}{2}\angle DHG \\ \Rightarrow \angle 1 = \angle 2{\text{ }}\left( {{\text{GM & HL are bisectors of }}\angle {\text{AGH and }}\angle {\text{DHG respectively}}} \right) \\ \Rightarrow GM\parallel HL{\text{ }}\left( {\angle 1{\text{ and }}\angle {\text{2 from a pair of alternate interior angles are equal}}} \right) \\ \]Similarly, \[GL\parallel MH\]So, GMHL is a parallelogram.Since \[AB\parallel CD\]We have \[\angle BGH + \angle DHG = {180^\circ}\] as the sum of interior angles on the same side of the transversal is equal to \[{180^\circ}\].Dividing both sides by 2, we have\[ \Rightarrow \dfrac{1}{2}\angle BGH + \dfrac{1}{2}\angle DHG = {90^\circ} \\ \Rightarrow \angle 3 + \angle 2 = {90^\circ}........................................................\left( 1 \right) \\ \]In \[\Delta GLH\] as the sum of the angles is equal to \[{180^\circ}\], we have\[ \Rightarrow \angle 2 + \angle 3 + \angle L = {180^\circ} \\ \Rightarrow {90^\circ} + \angle L = {180^\circ}{\text{ }}\left( {{\text{Using }}\left( 1 \right)} \right) \\ \Rightarrow \angle L = {180^\circ} - {90^\circ} \\ \therefore \angle L = {90^\circ} \\ \]Thus, in a parallelogram GMHL, \[\angle L = {90^\circ}\]We know that if one of the angles is the right angle in the parallelogram then it is also a rectangle.Hence, proved that GHML is a rectangle.Thus, if two parallel lines are intersected by a transversal, then bisectors of the interior angles form a rectangle.Note: The sum of interior angles on the same side of the transversal is equal to \[{180^\circ}\]. When two lines are crossed by another line which is called as the transversal the alternate interior angles are equal.

Sum of the two angles is 360o

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