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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;

add to each the angle ACB;

(Th. 9.)

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. I. 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.

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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 OROS.

Because in the triangles POR, POS

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.

(Hyp.)

(Hyp.)

(Th. 7.)

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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 OQR is the exterior angle of the triangle OQP;

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therefore ORQ is an acute angle, and less than OQR;

and therefore OR is greater than OQ.

(Th. 18.)

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.

Part. En. Let ABC, DEF be the two triangles, having the sides BA, AC equal to the sides ED, DF respectively, and having also the angle B = the angle E;

it is required to prove that the angle C is equal or supplementary to the angle F; and that when the angle C is equal to the angle F, the triangles are equal in all respects.

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Proof. The contained angle A must be either equal or unequal to the contained angle D.

If A = D, as in Fig. 1, then, by Theorem 5, the triangles are equal in all respects; and the angle C is equal to the angle F.

If A is not equal to D, as in Fig. 2, let the point A be placed on D, and AB along DE,

then, because AB = DE,

(Hyp.)

the point B will coincide with the point E;

and because the angle at B = the angle at E,

(Hyp.)

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and therefore the angle DFE = the angle DGF; (Th. 6.) but DGF is supplementary to DGE, that is, to ACB, and therefore the angle Fis supplementary to the angle C.

Q. E. D.

COR. Hence the triangles are equal in all respects(1) If the two angles given equal are right angles or obtuse angles.

For then the remaining angles must be acute, and therefore cannot be supplementary, and must therefore be equal by the Theorem, and therefore the triangles must be equal in all respects.

(2) If the angles opposite to the other two equal sides are both acute, or both obtuse, or if one of them is a right angle.

(3) If the side opposite the given angle in each triangle is not less than the other given side.

For then the given angles must be the greater of the two, and therefore the remaining angles must be both acute, and therefore cannot be supplementary, and must therefore be equal, by the Theorem, and therefore the triangles must be equal in all respects.

EXERCISES ON THEOREMS OF EQUALITY.

The general method to be adopted in the solution of theorems of equality is the following. Examine fully the statement of the question; see what is included among the data: what lines and angles are given equal by hypothesis. Then see what is required to be proved, what lines or angles have to be proved to be equal. It may follow from the properties proved of a single triangle; or it may depend on the equality of a pair of triangles. In the latter case examine the triangles of which they form corresponding parts, and see whether the data are sufficient to prove these triangles equal. If the data are sufficient, the solution is effected by comparing the triangles, and shewing the required equality of the lines and angles; if not, the data must be used to establish results, which in their turn can be used to establish the conclusion required.

The beginner will do well to arrange his proofs in the manner shewn in the example, giving references in the margin.

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