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For the point D being obviously the centre of a semicircle passing through G, F, and E, the angle GFE is a right angle; and the triangles EGF, EHI, having the sides GE, EF equal to HE, EI, and their contained angles vertical,—are equal (I. 3.), and consequently the angle HIE is equal to GFE, or is a right angle; but since CK and HI are parallel, the angle CKA is equal to HIE (I. 22.), and therefore is also a right angle, or CK is perpendicular to AB.


Geometrical Problems resolved by means of Compasses, or by the mere description of Circles.


To repeat a given distance in the same direction.

Let A and B be two given points; it is required to find, by means of compasses only, a series of equidistant points in the same extended line. - -

From B as a centre, with the given distance BA, describe a portion of a circle, in which inflect that distance three times to C ; from C, with the same radius, describe

another circle, and insert the triple chords to D; repeat that process from D, E, &c.: The equidistant points A, B, C, D, E, &c. will all lie in the same straight line. - For, by this construction, three equilateral triangles are formed about the point B, and consequently (I. 30. cor. 1.) the whole angle ABC, made by the opposite distances BA and BC, is equal to two right angles, or ABC is a straight line. The same reason applies to the successive points, D,

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To find the direction of a perpendicular from a given point to the straight line joining it with another given point.

Given the points A and B: to find a third point, such that the straight line connecting it with B shall be at right angles to BA. .

From A and B, with any convenient distance, describe two arcs intersecting in C, from which, with the same radius, describe a portion of a circle passing through the points A and B, and insert that radius three times from A to D: BD is perpendicular to BA.

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For it is evident, from the last Proposition, that the arc ABD is a semicircumference, and consequently (III. 19.) the angle ABD contained in it is a right angle.

Scholium. The construction would be somewhat simpli

fied, by taking the distance AB for the radius.


To find the direction of a perpendicular let fall from a given point upon the straight line which connects two given points.

Let C be a point, from which a perpendicular is to be let fall upon the straight line joining A and B. From A as a centre, with the

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distance AC, describe an arc, :

and from B as a centre, with the : distance BC, describe another ......... iX------------- -i--- - - - JB

arc, intersecting the former in the point D : CD is perpendicu- lar to AB. § For CAD and CBD are evidently isosceles triangles, and consequently (I. 7.) their vertices must lie in a straight line AB which bisects their base CD at right angles. - . Scholium. It will be perceived that this construction differs not in any respect from the mode employed in Prop. 6. Book I. of the Elements.


To bisect a given distance.

Let A and B be two given points; it is required to find the middle point in the same direction.

From B as a centre, with the radius BA, describe a semicircle, by inserting that distance successively from A to C, D, and E; from A as a centre, with the distance AE, describe a portion of a circle FEG, in which, from the point E, inflect the chords EF and EG equal to EC; and from the points F and G, with the same radius EC describe arcsintersecting in H : This point bisects the distance AB.

For, by the first Proposition, the points A, B, and E extend in a straight line; but the triangles FAG, FHG, and FEG, being evidently isosceles, their vertices A, H, and E (I. 7.) must - lie in a straight line; whence - Jó the point H lies in the direction AB. Again, because EFH is an isosceles triangle, AF*—HF*= EA.AH (II. 20.); that is, AE*—EC*, or (III. 19. and II. 10.) AC” or AB’=EA.AH. Wherefore, since EA is double of AB, the segment AH must be its half.

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To trisect a given distance.

Let it be required to find two intermediate points that are situate at equal intervals in the line of communication AB.

Repeat (App. II. 1.) the distance AB on both sides to C and D; from these points, with the radius CD, describe the arcs EDF and GCH, from D and C inflect the chords DE and DF, CG and CH, all equal to DB, IC and, with the same distance and from the points E and F, G and H, describe arcs intersecting in I and K: The distance AB is trisected by the points I and K.

For it may be demonstrated, as in the last proposition, that the points I and K lie in the same direction AB. In like manner, it appears (II. 20.) that DG"— KG*=CD.DK, or 9AB’—4AB°, or 5AB*=3AB.DK; and consequently 5AB=3DK, or 2AB=3AK, and AB= 3BK. But, for the same reason, AB=3AI.

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