when two straight lines intersect, they form congruent opposite (vertical) angles.

1. Construct line AB.[br]2. Construct line CD such that it intersects line AB.[br]3. Label the point of intersection of both lines point E. [br][br] We want to show i. m∠AEC=m∠DEB and[br]   ii. m∠AED=m∠BEC[br][br]i. Line AE cuts line CD, making the angles AEC and AED two right angles or equal to two right angles. [br]So, m∠AEC+m∠AED=180°. [I.13][br][br]Also, line DE cuts line AB making the angles AED and DEB two[br]right angles or equal to two right angles.[br] So, m∠DEB+m∠AED=180°. [I.13][br][br]So, if m∠AEC+m∠AED=180° and[br]  m∠DEB+m∠AED=180°, then[br][br]m∠AEC+m∠AED=m∠DEB+m∠AED (Post 4 and C.N 1)[br][br]Hence, m∠AEC=m∠DEB by subtraction and C.N 3.[br][br][br]ii. Similarly, we can show 2. m∠AED=m∠BEC. [br]Line BE cuts line CD making angles DEB and [br]CEB two right angles or equal to two right angles.[br]Aso, m∠DEB+m∠CEB=180° [1.13].[br][br]Also, line DE cuts line AB making the angles AED and DEB two right angles or equal to two right angles[br][br]So, m∠AED+m∠DEB=180° [1.13].[br][br]So, if m∠DEB+m∠CEB=180° and m∠AED+m∠DEB=180°, then [br][br]m∠DEB+m∠CEB=m∠AED+m∠DEB (Post 4 and C.N. 1)[br][br]Hence, m∠CEB=m∠AED by subtraction and C.N 3.[br][br]Q.E.D.