THEOREM XXXVII. In any triangle, the square of the side subtending an acute angle, is less than the squares of the base and the other side, by twice the rectangle of the base and the distance of the perpendicular from the acute angle. Let ABC be a triangle, having the angle A acute, and CD perpendicular to AB; then will the square of BC be less than the squares of AB, AC, by twice the rectangle of AB, AD. That is, BC2 AB2+ AC2 - 2 AD. AB. For, BD2 AD2 + AB2 — 2 AD. AB (th. 32). BDA And BD2+DC2 = AD2 + DC2 + AB2 — 2 AD. AB (ax. 2). THEOREM XXXVIII. In any triangle, the double of the square of a line drawn from the vertex to the middle of the base, together with double the square of the half base, is equal to the sum of the squares of the other two sides. Let ABC be a triangle, and CD the line drawn from the vertex to the middle of the base AB, bisecting it into the two equal parts AD, DB; then will the sum of the squares of AC, CB, be equal to twice the sum of the squares of CD, AD; or AC+ CB22 CD 2 + 2 AD*. For, AC2 = CD2 + AD2 + 2 AD. DE (th. 36). And, BC = CD2 + BD2 — 2 AD. DE (th. 37). Therefore, AC + BC2 = 2CD2 + AD2 + BD2 A = 2CD2 + 2AD2 (ax. 2). Q. E. D. THEOREM XXXIX. DEB In an isosceles triangle, the square of a line drawn from the vertex to any poins in the base, together with the rectangle of the segments of the base, is equal to the square of one of the equal sides of the triangle. Let ABC be the isosceles triangle, and CD a line drawn from the vertex to any point D in the base : then will the square of AC be equal to the square of CD, together with the rectangle of AD and DB. That is, AC CD + AD. DB. For AC. CD2 = AE2 DE2 (th. 35). - =AD. DB (th. 33). Therefore AC = CD2 + AD . DB (ax. 2). Q. E. D. THEOREM XL. In any parallelogram, the two diagonals bisect each other; and the sum of their squares is equal to the sum of the squares of all the four sides of the paral lelogram. Let ABCD be a parallelogram, whose diagonals intersect each other in E: then will AE be equal to EC, and BE to ED; and the sum of the squares of AC, BD, will be equal to the sum of the squares of AB, BC, CD, DA. That is, AE EC, and BE = ED, and ACBD = AB2 + BC2 + CD2 + DA2, E For, the triangles AEB, DEC, are equiangular, because they have the opposite angles at E equal (th. 7), and the two lines AC, BD, meeting the parallels AB, DC, make the angle BAE equal to the angle DCE, and the angle ABE equal to the angle CDE, and the side AB equal to the side DC (th. 22); therefore, these two triangles are identical, and have their corresponding sides equal (th. 2), viz. AE= EC, and BE=ED. Again, since AC is bisected in E, the sum of the squares AD2 + DC2 = 2AE' +2DE (th. 38). In like manner, AB2 + BC2 = 2AE2 + 2BE2 or 2DE2. Therefore, AB2 + BC2 + CD2 + DA2 = 4AE2 + 4DE2 (ax. 2). But, because the square of the line (cor. th. 31), that is, Therefore, AB2 + BC2 + a whole line is equal to 4 times the square of halt AC2 = 4AE*, and BD2 = 4DE2: CD2 + DA2 = AC2 + BD2 (ax. 1). Q. E. D. Cor. 1. If AD = DC, or the parallelogram be a rhombus; then AD2 = AŁ1 + ED2, CD2= DE2 + CE*, &c. Cor. 2. Hence, and by th. 34, the diagonals of a rhombus intersect at right angles. THEOREM XLI. If a line, drawn through or from the centre of a circle, bisect a chord, it wik be perpendicular to it; or, if it be perpendicular to the chord, it will bisect both the chord and the arc of the chord. Let AB be any chord in a circle, and CD a line drawn from the centre C to the chord. Then, if the chord be bi ected in the point D, CD will be perpendicular to AB. B Draw the two radii CA, CB. Then the two triangles ACD, BCD, having CA equal to CB (def. 44), and CD common, also AD equal to DB (by hyp.); they have all the three sides of the one, equal to all the three sides of the other, and so have their angles also equal (th. 5). Hence, then, the ange ADC being equal to the angle BDC, these angles are right angles, and the line CD is perpendicular to AB (def. 11). Again, if CD be perpendicular to AB, then will the chord AB be bisected at the point D, or have AD equal to DB; and the arc AEB bisected in the point E, or have AE equal EB. For, having drawn CA, CB, as before; then, in the triangle ABC, because the side CA is equal to the side CB, their opposite angles A and B are also equal (th. 3). Hence, then, in the two triangles ACD, BCD, the angle A is equal to the angle B, and the angles at D are equal (def. 11); therefore, their third angles are also equal (corol. 1, th. 17). And having the side CD) common, they have also the side equal to the side DB (th. 2). Also, since the angle ACE is equal to the angle BCE, the arc AE, which measures the former (def. 57), is equal to the arc BE, which measures the lat ter, since equal angles must have equal measures. Corol. Hence, a line bisecting any chord at right angles, passes through the centre of the circle. THEOREM XLII. If more than two equal lines can be drawn from any point within a circle the circumference, that point will be the centre. Let ABC be a circle, and D a point within it: then, if any three lines, DA, DB, DC, drawn from the point D to the circumference, be equal to each other, the point D will be the centre. Draw the chords AB, BC, which let be bisected in the points E, F, and join DE, DF. Then, the two triangles DAE, DEE, have the side DA equal to the side DB by supposition, and the side E B AE equal to the side EB by hypothesis, also the side DE common: therefore, these two triangles are identical, and have the angles at E equal to each other (th. 5); consequently, DE is perpendicular to the middle of the chord AB (def. 11), and therefore passes through the centre of the circle (corol. th. 41). In like manner, it may be shown that DF passes through the centre. Consequently, the point D is the centre of the circle, and the three equal lines DA, DB, DC, are radii. Q. E. D. THEOREM XLIII. If two circles, placed one within another, touch, the centres of the circles and the point of contact will be all in the same right line. Let the two circles ABC, ADE, touch one another internally in the point A; then will the point A and the centres of those circles be all in the same right line. Let F be the centre of the circle ABC, through which draw the diameter AFC. Then, if the centre of the other circle can be out of this line AC, let it be supposed in some other point as G; through which draw the line FG, cutting the two circles in B and D. B F G E Now, in the triangle AFG, the sum of the two sides FG, GA, is greater than the third side A (th. 10), or greater than its equal radius FB. From each of these take away the common part FG, and the remainder GA will be greater than the remainder GB. But the point G being supposed the centre of the inner circle, its two radii, GA, GD, are equal to each other; consequently, GD will also be greater than GB. But ADE being the inner circle, GD is necessarily less than GB. So that GD is both greater and less than GB; which is absurd. To get quit of this absurdity we must abandon the supposition that produced it, which was that G might be out of the line AFC. Consequently, the centre G cannot be out of the line AFC. Q. E. D. THEOREM XLIV. If two circles touch one another externally, the centres of the circles and point of contact will be all in the same right line. Let the two circles ABC, ADE, touch one another externally at the point A; then will the point of contact A and the centres of the two circles be all in the same right line. Let F be the centre of the circle ABC, through which draw the diameter AFC, and produce it to the other circle at E. Then, if the centre of the other circle ADE can be out of the line FE, let it, if possible, be supposed in some other point as G; and draw the lines AG, FBDG, cutting the two circles in B and D. D Then, in the triangle AFG, the sum of the two sides AF, AG, is greater than the third side FG (th. 10). But, F and G being the centres of the two circles, the two radii GA, GD, are equal, as are also the two radii AF, fb. Hence the sum of GA, AF, is equal to the sum of GD, BF; and, therefore, this latter sum also, GD, BF, is greater than GF, which is absurd. Consequently, the centre G cannot be out of the line EF. Q. E. D. THEOREM XLV. Any chords in a circle, which are equally distant from the centre, are equal to each other; or if they be equal to each other, they will be equally distant from the centre. Let AB, CD, be any two chords at equal distances from the centre G; then will these two chords AB, CD, be equal to each other. Draw the two radii GA, GC, and the two perpendi. culars GE, GF, which are the equal distances from the centre G. Then, the two right-angled triangles, GAE, GCF, having the side GA equal the side GC, and the side GE equal the side GF, and the angle at E equal to the angle at F, therefore those two triangles are identical (cor. 2, th. 34), and have the line AE equal to the line CF. But AB is the double of AE, and CD is the double of CF (th. 41); therefore AB is equal to CD (by ax. 6). Q. E. D. Again, if the chord AB be equal to the chord CD; then will their distances from the centre, GE, GF, also be equal to each other. For, since AB is equal CD by supposition, the half AE is equal the half CF. Also, the radii GA, GC, being equal, as well as the right angles E and F, therefore the third sides are equal (cor. 2, th. 34), or the distance GE equal the distance GF. Q. E. D. THEOREM XLVI. A line perpendicular to the extremity of a radius, is a tangent to the circle. Let the line ADB be perpendicular to the radius CD of a circle; then shall AB touch the circle in the point D only. From any other point E in the line AB draw CFE to the centre, cutting the circle in F. Then, because the angle D, of the triangle CDE, is a right angle, the angle at E is acute (cor. 3, th. 17), and consequently less than the angle D. But the greater side D E B is always opposite to the greater angle (th. 9); therefore the side CE is greater than the side CD, or greater than its equal CF. Hence the point E is without the circle; and the same for every other point in the line AB. Consequently the whole line is without the circle, and meets it in the point D only. THEOREM XLVII. When a line is a tangent to a circle, a radius drawn to the point of contact is perpendicular to the tangent. Let the line AB touch the circumference of a circle at the point D; then will the radius CD be perpendicular to the tangent AB. [See the last figure.] For, the line AB being wholly without the circumference except at the point D, every other line, as CE, drawn from the centre C to the line AB, must pass out of the circle to arrive at this line. The line CD is therefore the shortest that can be drawn from the point C to the line AB, and consequently (th. 21,) it is perpendicular to that line. Corol. Hence, conversely, a line drawn perpendicular to a tangent, at the point of contact, passes through the centre of the circle. THEOREM XLVIII. The angle formed by a tangent and chord is measured by half the arc of that chord. Let AB be a tangent to a circle, and CD a chord drawn from the point of contact C; then is the angle BCD measured by half the arc CDF, and the angle ACD measured by half the arc CGD. Draw the radius EC to the point of contact, and the radius EF perpendicular to the chord at H. Then the radius EF, being perpendicular to the chord CD, bisects the arc CFD (th. 41). Therefore CF is half the arc CFD. In the triangle CEH, the angle H being a right one, the sum of the two remaining angles E and C is equal to a right angle (cor. 3, th. 17), which is equal to the angle BCE, because the radius CE is perpendicular to the tan A B F gent. From each of these equals take away the common part or angle C, and there remains the angle E equal to the angle BCD. But the angle E is measured by the arc CF (def. 57), which is the half of CFD; therefore the equal angle BCD must also have the same measure, namely, half the arc CFD of the chord CD. Again, the line GEF, being perpendicular to the chord CD, bisects the arc CGD (th. 41). Therefore CG is half the arc CGD. Now, since the line CE, meeting FG, makes the sum of the two angles at E equal to two right angles (th. 6), and the line CD makes with AB the sum of the two angles at C equal to two right angles; if from these two equal sums there be taken away the parts or angles CEH and BCH, which have been proved equal, there remains the angle CEG equal to the angle ACH. But the former of these, CEG, being an angle at the centre, is measured by the arc CG (def. 57); consequently the equal angle ACD must also have the same measure CG, which is half the arc CGD of the chord CD. Q. E. D. Corol. 1. The sum of the two right angles is measured by half the circumference. For the two angles BCD, ACD, which make up two right angles, are |