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PRO P. III. THEOREM.

If a right line be perpendicular to two other right lines, at their point of interfection, it will also be perpendicular to the plane which paffes through thofe lines.

B

Let the right line AB be perpendicular to each of the two right lines BC, BD, at their point of interfection B ; then will it also be perpendicular to the plane which paffes through thofe lines.

For make BD equal to BC; and, in the plane which paffes through those lines, draw any right line BE; and join the points CD, AD, AE and AC:

Then because the fide BC is equal to the fide BD (by Conft.), and the perpendicular AB is common to each of the triangles ABC, ABD, the fide AD will also be equal to the fide AC (1. 4.)

And fince the triangles CAD, CBD are isofceles, the rectangle of CE, ED, together with the square of EB, is equal to the fquare of DB; and the rectangle of CE, ED together with the fquare of EA, is equal to the fquare of AD (II. 20.)

From each of these equals, take away the rectangle of CE, ED which is common, and the difference of the

fquares

be equal to the rectangle of AD, DB, together with the fquare of CD.

For, about the triangle ABC, describe the circle AEC (IV. 5.), cutting CD, produced, in E; and join EB.

Then, because the angle ACD is equal to the angle ECB (by Hyp.), and the angle CAD to the angle CEB (III. 15.), the remaining angle ADC will be equal to the remaining angle CBE (I. 28. Cor.)

The triangles CAD, CEB being, therefore, equiangular, CA will be to CD as CE to CB (VI. 5.); and confequently the rectangle of CA, CB is equal to the rectangle of CE, CD (VI. 12.)

But the rectangle of CE, CD is equal to the rectangle of ED, DC, together with the fquare of CD (II. 10.); whence the rectangle of CA, CB is alfo equal to the rectangle of ED, DC, together with the fquare of CD.

And fince the rectangle of ED,

angle of AD, DB (III. 27.), the

DC is equal to the rect

rectangle of AC, CB is

alfo equal to the rectangle of AD, DB, together with the

fquare of CD.

Q. E. D.

PRO P. XXVI. THEOREM..

The rectangle of the two fides of any triangle, is equal to the rectangle of the perpendicular, drawn from the vertical angle to the bafe, and the diameter of the circumfcribing circle.

Let ABC be a triangle, having CD perpendicular to AB; then will the rectangle of AC, CB be equal to the rectangle of CD and the diameter of the circumfcribing circle.

For, about the triangle ABC, defcribe the circle AEC (IV. 5.); in which draw the diameter CE; and join EB.

Then, fince the angle CAD is equal to the angle CEB (III. 15.) and the angle ADC to the angle EBC, being each of them right angles (Conft. and III, 16.), the remaining angle ACD will be equal to the remaining angle ICB (I. 28. Cor.)

The triangles ACD, ECB are, therefore, equiangular; whence AC is to CD as CE is to CB (VI. 5.); and confequently the rectangle of AC, CB is equal to the rectangle of CD, CE (VI. 12.) Q. E. D. SCHOLIUM. When ABC is an obtufe angle, the perpendicular CD falls without the circle; but the fame demonstration will hold.

PROP.

PRO P. XXVII.

THEOREM.

The rectangle of the two diagonals of any quadrilateral, infcribed in a circle, is equal to the fum of the rectangles of its oppofite fides.

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Let ABCD be any quadrilateral infcribed in a circle, of which the diagonals are AC, BD; then will the rectangle of AC, BD be equal to the rectangles of AB, DC and

AD, BC.

For make the angle CDE equal to the angle ADB (I. 20.); then, if to each of thefe angles, there be added the common angle EDB, the angle ADE will be equal to the angle CDB.

The angle DAE is alfo equal to the angle DBC, being angles in the fame fegment, whence the remaining angle AED is equal to the remaining angle BCD (I. 28. Cor.)

Since, therefore, the triangles ADE, BDC are equiangular, AD is to AE as BD is to BC (VI. 5.); and confequently the rectangle of AD, BC is equal to the rectangle of AE, BD (VI. 12.)

Again, the angle CDE being equal to the angle ADB (by Conft.), and the angle ECD to the angle ABD (III. 15.), the remaining angle CED will be equal to the remaining angle BAD (1. 28. Cor.)

The triangles CED, ADB are, therefore, alfo equiangular; whence AB is to BD as EC is to DC (VI. 5.); and confequently the rectangle of AB, DC is equal to the rectangle of EC, BD. (VI. 12.)

And if, to these equals, there be added the former, the rectangle of AB, DC together with the rectangle of ad, BC will be equal to the rectangle of EC, BD together with the rectangle of AE, BD.

But the rectangles of AE, BD, and EC, BD are equal to the rectangle of AC, BD (II. 8.); whence the rectangle of AC, BD is also equal to the rectangles of AB, DC and AD, BC.

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