T16 If a line in a plane is perpendicular to one of two parallel lines in the same plane, then it is also perpendicular to the other.

Idea of Proof . Follows from T15 and the parallel axiom. (PA)

T17 If two lines are cut by a transversal so that the alternate interior angles are congruent, then the lines are parallel.

Idea of Proof. Assume they meet. Get a violation of the exterior angle theorem (T11).

T18 If two parallel lines are cut by a transversal, then the alternate interior angles are congruent.

Restatement . Given lines l and m cut by a transversal t at A and B forming alternate interior angles <1 and <2 shown in diagram. Show <1 is congruent with <2.

Idea of Proof. Use AAA to construct line l' thru A so that the alternate interior angles formed are congruent. Then by T17 the line l' is parallel with m. By PA, l' = l. Hence <1 is congruent with <2. qed.

T19 The sum of the measures of the angles of a triangle is 180 degrees.

Restatement . Given ABC Show m( angle (A)) + m(angle(B)) + m(angle(C)) = 180.

Idea of Proof . Use AAA to construct a line l thru C which is parallel with line(AB). angle(A) and angle(1) are congruent (T18). angle(B) and angle(2) are congruent. (T18). angle(1) + angle(C) + angle(2) = 180. qed.

T20 The sum of the measures of the interior angles of a convex n-gon is (n-2)180 degrees.

Idea of Proof. Pick a point in the interior of the n-gon and dissect the n-gon into n triangles, each of which, by T19, has 180 in it, so the sum of the interior angles is degrees. qed

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T21 The sum of the measures of the exterior angles of a convex n-gon, one at each vertex, is 360 degrees.

Idea of Proof . The sum of those exterior angles together with their associated interior angles is degrees, and this is also by T20 + sum of exterior angles. From this it follows that sum of exterior angles is 360. qed .

T22 A diagonal of a parallelogram divides it into two congruent triangles and hence the opposite sides of a parallelogram are congruent.

Idea of Proof. Use T18 twice and ASA to prove the first part. The second part follows from the first part.

T23 If the opposite angles are congruent and the adjacent angles are supplementary or the opposite sides of a quadrilateral are congruent, the quadrilateral is a parallelogram.

Idea of Proof. In the first case, use T17 twice to get it. In the second case use SSS and then T17 twice to get it.

T24 The diagonals of a quadrilateral bisect each other if and only if the quadrilateral is a parallelogram.

Given : quadrilateral ABCD. To show : seg(AC) and seg(BD) intersect at the midpoint of each if and only if AB is parallel to CD and AD is parallel to BC.

Idea of Proof . if part: Let M be the point of intersection. By T22 AD = CB. By T18, angle(ADM) is congruent with angle MBC and angle(DAM) is congruent with angle(BCM). So by ASA, triangle(DAM) is congruent with triangle(BCM). Hence the corresponding sides CM = MA and DM = MB.

only if part : Let M be the point of intersection. By the vertical angle theorem T7, angle(DMA) is congruent with angle(BMC). Since the diagonals bisect each other, DM = BM and CM = AM. Hence triangle(DAM) is congruent with triangle(BCM) by SAS. But then angle(DAM) is congruent with angle(BCM), and so line(AD) is parallel with line(BC) by T17. In the same manner, show line DC is paralle with line(AB). qed

T25 Two distinct lines parallel to a third are parallel to each other.

Idea of Proof. Get this straight from the parallel axiom PA.

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