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PRO P. XI. PROBLEM.

At a given point, in a given right line, to erect a perpendicular.

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Let as be the given right line, and the given point in it; it is required to draw a right line, from the point D, that shall be perpendicular to AB,

Take any point e, in AB, and make of equal to DE (Prop. 3.), and upon EF describe the equilateral triangle ECF (Prop. 1.)

Join the points d, c; and the right line cd will be perpendicular to AB, as was required.

For ce is equal to CF (Def. 16), ED to DF (by Conft.) and cd is common to each of the triangles ECD, FCD.

The three sides of the triangle Ecd being, therefore, equal to the three sides of the triangle FCD, each to each, the angle Edc will, also, be equal to the angle FDC (Prop. 7.)

But one right line is perpendicular to another when the angles on both sides of it are equal (Def. 8.); therefore cd is perpendicular to AB ; and it is drawn from the point p as was to be done.

SCHOLIUM. If the given point be at, or near, the end of AB, the line must be produced.

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To draw a light line perpendicular to a given right line, of an unlimited length, from a given point without it.

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Let AB be the given right line, and c the given point; it is required to draw a right line from the point c, that shall be perpendicular to AB.

Take any point D, in AB, and from that point, with the distance Dc, describe the circle cge, cutting AB

in G.

Join gc, and from the point G, with the distance Gc, describe the circle n E m, cutting the former in E.

Through the points C, E draw the right line CFE, cut* ting AB in F, and cf will be perpendicular to AB, as was required.

For, join the points D,C, D, E, and 'G,E :

Then, because pc is equal to DE, GC to Ge (Def. 13.) and DG common to each of the triangles DCG, DEG, the angle CDG will be equal to the angle GDE (Prop. 7.)

And, since Dc is equal to DE, DF common to each of the triangles DCF, DEF, and the angle CDG equal to the angle gde, the angle DFC will also be equal to the angle DFE (Prop. 4.)

But one line is perpendicular to another when the angles on both sides of it are equal (Def. 8.); therefore cf is perpendicular to AB; and it is drawn from the point c, as was to be done.

PRO P. XIII. THEOREM.

The angles which one right line makes with another, on the fame side of it, are together equal to two right angles.

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Let the right line AB fall upon the right line CD; then will the angles ABC, ABD, taken together, be equal to two right angles.

For if the angles ABC, ABD be equal to each other, they will be, each of them, right angles (Def. 8 and 9.)

But if they be unequal, let El be drawn, from the point B, perpendicular to CD (Prop. II.)

Then, fince the angles EBC, EBD are right angles (Def. 8.), and the angle EBD is equal to the angles EBA, ABD (Ax. 8.), the angles EBC, EBA and ABD will be equal to two right angles.

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But the angles EBC, EBA are, together, equal to the angle ABC (Ax. 8.); confequently the angles ABC, ABD are, also, equal to two right angles. Q. E. D.

COROLL. All the angles which can be made, at any point B, on the same side of the right line cd, are, together, equal to two right angles.

PROP. xiv. Theorem.

If a right line meet two other right lines, in the same point, and make the angles on each side of it together equal to two right angles, those lineś will form one continued right line.

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Let the right line AB meet the two right lines CB, BD, at the point B, and make the angles ABC, ABD together equal to two right angles, then will be be in the same right line with co.

For, if it be not, let some other line Be be in the same right line with CB.

Then, because the right line AB falls upon the right line cbe, the angles ABC, ABE, taken together, are equal to two right angles (Prop. 13.)

But the angles ABC, ABD are also equal to two right angles (by Hyp.); consequently the angles ABC, ABE are equal to the angles ABC, ABD.

And, if the angle ABC, which is common, be taken away, the remaining angle abe will be equal to the remaining angle ABD; the less to the greater, which is absurd.

The line be, therefore, is not in the same right line with CB; and the same may be proved of any other line but BD; confequently cBD is one continued right line, as was to be shewn.

PRO P. XV. THEO R E M.

If two right lines intersect each other, the opposite angles will be equal,

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Let the two right lines AB, co interfect each other in the point E; then will the angle Abc be equal to the angle des, and the angle Aed to the angle ceB.

For, since the right line ce falls upon the right line AB, the angles aec, CEB, taken together, are equal to two right angles (Prop. 13.)

And, because the right line be falls upon the right line cd, the angles BED, CEB, taken together, are also equal to two right angles (Prop. 13.)

The angles AEC, Ceb, taken together, are, therefore, equal to the angles bed, ces taken together (Ax. I.)

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