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Book VI. Cor. The triangle ADE is similar to the triangle ABC.

For the two triangles BAD, CAE having the angles at D and E right angles, and the angle at A common, are equiangular, and therefore BA : AD:: CA: AE, and alternately BA:CA:: AD: AE; therefore the two triangles

BAC, DAE, have the angle at A common, and the sides h 6.6. about that angle proportionals, therefore they are equiangular h

and similar.

Hence the rectangles BA.AE, CA.AD are equal.

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F from any angle of a triangle a perpendicular be

drawn to the opposite side, or base; the rectangle contained by the sum and difference of the other two sides, is equal to the rectangle contained by the sum and difference of the segments, into which the base is divided by the perpendicular.

Let ABC be a triangle, AD a perpendicular drawn from the angle A on the base BC, so that BD, DC are the segments of the base ; AC+AB. AC-AB-CD+DB. CD-DB.

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From A as a centre with the radius AC, the greater of the two fides, describe the circle CFG; produce AB to meet the circumference in E and F, and CB to meet it in G. Then because AFSAC, BF=AB+AC, the sum of the fides; and Since AESAC, BESAC-AB - the difference of the fides.

Also,

Also, because AD drawn from the centre cuts GC at right Book VI. angles, it bisects it; therefore, when the perpendicular falls within the triangle, BG=DG-DB=DC-DB = the difference of the segments of the base, and BC=BD+DC = the sum of the segments. But when AD falls without the triangle, BG=DG+DB=CD+DB = the sum of the segments of the base, and BC=CD-DB = the difference of the fegments of the base. Now, in both cases, because B is the intersection of the two lines FE, GC, drawn in the circle, FB.BE=CB.BG; that is, as has been shown, AC+AB. AC-AB=CD+DB. CD-DB. Therefore, &c. Q. E. D.

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