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PROPOSITION XII. THEOREM.

69. CONVERSELY: When two straight lines are cut by a third straight line, if the alternate-interior angles be equal, the two straight lines are parallel.

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Let E F cut the straight lines A B and C D in the points

H and K, and let the Z A H K = 2 H K D.

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then

Through the point H draw M N || to CD;

Z MH K = H K D,

(being alt.-int. £). LAHK= Z HKD,

§ 68

But

Hyp.

... MH K= LA H K.

Ax. 1.

.. the lines M N and A B coincide.

Cons.

But

M N is II to CD;
.. A B, which coincides with M N, is |to C D.

Q. E, D.

PROPOSITION XIII. THEOREM.

70. If two parallel lines be cut by a third straight line, the exterior-interior angles are equal.

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Let A B and C D be two parallel lines cut by the

straight line E F, in the points H and K.

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71. COROLLARY. The alternate-exterior angles, E H B and CKF, and also A H E and DK F, are equal.

PROPOSITION XIV. THEOREM.

72. CONVERSELY: When two straight lines are cut by a third straight line, if the exterior-interior angles be equal, these two straight lines are parallel.

Let EF cut the straight lines A B and C D in the

points II and K, and let the Z EHB= {HKD.

We are to prove A B || to CD.
Through the point H draw the straight line M N 1 to C D.
Then
ZEHN= L H K D,

§ 70
(being ext.-int. {).
Z EHB= L H K D.
..Z EHB= L EHN.

Ax. 1.

But

Hyp.

... the lines M N and A B coincide.

But

M N is || to CD,

Cons.

... A B, which coincides with M N, is il to C D.

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PROPOSITION XV. THEOREM.

73. If two parallel lines be cut by a third straight line, the sum of the two interior angles on the same side of the secant line is equal to two right angles.

Let A B and C D be two parallel lines cut by the

straight line E F in the points H and K.

We are to prove <BHK + Z HKD = two rt. .

ZEHB + 2BHK = 2 rt. L, $ 34

(being sup.-adj. 6). ZEHB=2 HKD,

$ 70 (being ext. -int. £). Substitute 2 HKD for 2 EH B in the first equality ;

But

2 BHK + ZHKD = 2 rt. I.

then

Q. E. D.

Proposition XVI. THEOREM. 74. CONVERSELY: When two straight lines are cut by a third straight line, if the two interior angles on the same side of the secant line: be together equal to two right angles, then the two straight lines are parallel.

Let EF cut the straight lines A B and C D in the

points H and K, and let the < B II K + 2 HKD equal two right angles.

We are to prove AB || to C D.
Through the point H draw M N || to C D.
Then NHK + Z HKD = 2 rt. 4, $73

(being two interior & on the same side of the secant line).
But Z BII K+HKD = 2 rt. &. Hyp.
.:. ZNH K + 2 HKD=ZBI K + 2 H K D. Ax. 1.
Take away from each of these equals the common ZH KD,
then 2 NHK= Z BH K.

.. the lines A B and M N coincide.
But
M N is || to CD;

Cons. .. A B, which coincides with MN, is II to C D.

Q. ED.

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