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Part. En. Let BAC, DEF be two triangles which

BA=DE,

AC = EF, but the base BC greater than the base DF;)

it is required to prove that the angle BAC is greater than the angle DEF.

Proof. For the angle BAC must either be equal to the angle DEF, or less than the angle DEF, or greater than the angle DEF.

But the angle BAC is not equal to the angle DEF, : for then the base BC would be equal to the base DF; but it is not

(Th. 5.) Nor is the angle BAC less than the angle DEF, for then the base BC would be less than the base DF; but it is not.

(Th. 14.) Therefore the angle BAC must be greater than the angle DEF*.

Q. E. D.

THEORER woles of the the siriangles.asite

If two triangles have two angles of the one equal to two angles of the other, each to each, and have the sides opposite to one of the equal angles in each equal, then the triangles are equal in all respects, those sides being equal which are opposite to the equal angles.

* Euclid I. 25.

Part. En. Let the two triangles ABC, DEF have the two angles

ABC=DEF,

ACB=DFE,
and the side AB=the side DE;)

it is required to prove the triangles are equal in all respects.

Proof. Let the point A be placed on the point D, and AB along DE, then because AB = DE,

(Hyp.) therefore the point B will fall on the point E.

And because the angle ABC is equal to the angle DEF,

(Hyp.) therefore the line BC will lie along EF. And the point C will fall on F, for if it fell otherwise as G, then, since the angle ACB is equal to the angle EFD, (Hyp.) the angle EGD would be equal to the angle EFD, the exterior angle equal to the interior and opposite, which is impossible ;

(Th. 9.) therefore the triangles would coincide and are equal in all respects, AC being equal to DF, BC to EF, and the angle BAC to the angle EDF*.

Q. E. D. * Eucl. 1. 26. Part 2.

THEOREM 18.

Any two angles of a triangle are together less than two right angles.

Part. En. Let ABC be a triangle, it is required to prove that any two of its angles ABC and ACB are together less than two right angles.

Proof. Produce the side BC to D.

Then because the exterior angle ACD is greater than the interior and opposite angle ABC;

(Th. 9.) add to each the angle ACB; therefore the two angles ACD and ACB are greater than the two ABC and ACB.

But ACD and ACB are together equal to two right angles;

(Th. 2.) therefore ABC and ACB are together less than two right angles *.

Q. E. D.

Cor. 1. If one angle of a triangle is right or obtuse, the others are acute.

COR. 2. From a given point outside a given straight line, only one perpendicular can be drawn to that straight line.

* Euclid, 1. 17.

THEOREM 19. Of all the straight lines that can be drawn from a given point to meet a given straight line, the perpendicular is the shortest; and of the others, those making equal angles with the perpendicular are equal; and that which makes a greater angle with the perpendicular is greater than that which makes a less.

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Part. En. Let O be the given point, and AB the given straight line, and let OP be the perpendicular, OQ an oblique; it is required to prove first that OP is less than OQ.

Proof. Since any two angles of a triangle are together less than two right angles;

(Th. 18.) therefore OPQ and OQP are together less than two right angles : but OPQ is a right angle;

(Hyp.) therefore OQP is less than a right angle.

And in the triangle OQP, since the angle OPQ is greater than the angle OQP; therefore OQ is greater than OP.

(Th. 11.) Again, let OS, OR be obliques making equal angles with the perpendicular OP; it is required to prove that OR=OS.

Because in the triangles POR, POS

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the angle OPR= OPS, being right angles,) and the angle POR = POS,

and PO is common; therefore the triangles are equal in all respects, and therefore OR = OS.

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Lastly, let OR make a greater angle with the perpendicular than OQ; it is required to prove that OR is greater than OQ.

Because O QR is the exterior angle of the triangle OQP; therefore QR is greater than OPQ; (Th. 9.) but : OPQ is a right angle; (Hyp.) therefore OQR is an obtuse angle ; therefore ORQ is an acute angle, and less than OQR;

(Th. 18.) and therefore OR is greater than OQ.

Q. E. D. Cor. Not more than two equal straight lines can be drawn from a given point to a given straight line.

THEOREM 20. If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles opposite to two equal sides equal, the angles opposite to the other two equal sides are either equal or supplementary, and in the former case the triangles are equal in all respects.

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