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Draw the perpendicular AD. Then, because AD is drawn from the right angle perpendicular to the hypotenuse, the triangles ABD, ADC are similar (VI. 8.) to the whole triangle ABC, and to one another; and therefore (VI. 4.) as CB : BA :: BA: BD; and (VI. 20. cor. 2.) because these three straight lines are proportionals, as CB : BD :: E:F: and inversely, (V. B.) as DB : BC ::F: E. For the same reason, as
DC: BC :: G: E. Wherefore (V. 24.) as B D and DC together are to BC, so are the figures F and G together to the figure E. But BD and DC together are equal to BC: therefore (V. A.) the figure E is equal to the figures F and G. Wherefore, the rectilineal figure, &c.
Schol. The 47th proposition of the first book is the case of this proposition, in which the figures described on the three sides are squares.
PROP. XXXII. THEOR. If two triangles which have two sides of the one proportional to two sides of the other, be joined at one angle, so as to have their homologous sides parallel to one another, and the angles contained by those sides, opening towards the same parts ; the remaining sides are in a straight line.
Let ABC, DCE be two triangles which have the two sides BA, AC proportional to the two CD, DE, viz., BA : AC:: CD: DE; and let AB be parallel to DC, and AC to DE, the angles A and D opening in the same direction ; BC, CE are in a straight line.
Because AB is parallel to DC, and AC meets them, the alternate angles BAC, ACD are (I. 29.) equal : for the same reason, the angle CDE is equal to ACD: wherefore also (I. ax. 1.) BAC is equal to CDE. And because the triangles ABC, DCE have the angles A and D equal, and the sides about these angles proportionals, viz., BA: AC::CD: DE, these triangles are (VI.6.) equiangular. Therefore the angles ABC, DCE are equal: and the angle BAC was proved to be equal to ACD. Therefore the whole ACE is equal to the two ABC, BAC: add ACB; then the angles ACE, ACB are equal to the angles ABC, BAC, ACB; that is, (I. 32.) to two right angles. Now, since at the point C, in A Ċ, the two lines BC, CE, which are on the opposite sides of it, make the adjacent angles ACE, ACB equal to two right angles; therefore (I. 14.) BC and CE are in a straight line. If, therefore, two triangles, &c.
PROP. XXXIII. THEOR. In equal circles, or in the same circle, angles, whether at the centres or circumferences, have the same ratio, as the arcs on which they stand, have to one another : so also have the sectors.
Let ABC, DEF be equal circles; and let BGC, EHF be angles at their centres, and BAC, EDF, be angles at their circumferences ; as the arc BC to the arc EF, so is the angle BGC to EHF, and the angle BAC to EDF; and also the sector BGC to the sector EHF.
Take any number of arcs CK, KL, each equal to BC, and any number FM, MN, each equal to EF: and join GK, GL, HM, HN. Because the arcs BC, CK, KL are all equal, the angles BGC, CGK, KGL are also (III. 27.) all equal : therefore whatever multiple the arc BL is of BC, the same multiple is the angle BGL of BGC. For the same reason, whatever multiple the arc EN is of EF, the same multiple is the angle EHN of EHF. Also, if the arc BL be equal to EN, the angle BGL is also equal
(III. 27.) to EHN; and if BL be greater than EN, likewise the angle BGL is greater than EHN; and if less, less. Therefore (V. def. 5.) the arc BC is to the arc EF, as the angle BGC to EHF. But (V. 15.) the angle BGC is to EHF, as the angle BAC to EDF, for (III. 20.) each is double of each : therefore, as the arc BC is to EF, so is the angle BGC to EHF, and the angle BAC to EDF.
Also, as the arc BC is to EF, so is the sector BGC to the sector EHF. In the arcs BC, CK, take any points X, O, and join BC, CK, BX, XC, CO, OK. Then, because in the triangles GBC, GCK, the sides BG, GC are equal to CG, GK, and that they contain equal angles, BC is equal (I. 4.) to CK, and the triangle GBC to GCK. Again, because the arc BC is equal to the arc CK, the remaining part BAC of the whole circumference, is equal to the remaining part CAK: wherefore (III. 27.) the angles BXC, COK, are equal ; and the segment BXC is therefore similar (III. def. 8.) to the segment COK; and they are upon equal straight lines BC, CK: they are therefore equal (III. 24.) to one another. And since the triangles BGC, CGK are equal, therefore (I. ax. 2.) the whole, the sector BGC, is equal to the whole, the sector CGK. For the same reason, the
sector KGL is equal to each of the sectors BGC, CGK. In like manner, the sectors EHF, FHM, MHN may be proved equal to one another. Therefore, whatever multiple the arc BL is of BC, the same multiple is the sector BGL of BGC; and whatever multiple the arc EN is of EF, the same multiple is the sector
EHN of EHF: and if the arc BL be equal to EN, the sector BGL is equal to EHN; if the arc BL be greater than EN, the sector BGL is greater than EHN; and if less, less. Therefore, (V. def. 5.) as the arc BC is to EF, so is the sector BGC to EHF. Wherefore, in equal circles, &c.
PROP. B. THEOR.
If an angle of a triangle be bisected by a straight line which likewise cuts the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the seginents of the base, together with the square of the straight line bisecting the angle.
Let ABC be a triangle, and let the angle BAC be bisected by AD; the rectangle BA.AC is equal to the rectangle BD.DČ, together with the square of AD.
Describe the circle (IV. 5.) ACB about the triangle ; produce AD to meet the circumference in E; and join EC. Then (hyp.) the angles BAD, CAE are equal; as are also (III. 21.) the angles B and E, for they are in the same segment: therefore (VI. 4. cor.) in the triangles ABD, AEC, as BA: AD:: EA : AC; and consequently (VI. 16.) the rectangle BA.AC is equal to EA. AD, that is (II. 3.) ED.DA, together with the square of AD. But (III. 35.) the rectangle ED.DA is equal to the rectangle BD.DC: therefore the rectangle BA.AC is equal to BD.DC, together with the square of AD:* wherefore, if an angle, &c.
* From this proposition in connexion with the 3d proposition of this book, we have the means of computing AD, when the sides are given in numbers. For, by the 3d, BA : AC :: BD: DC; and, by composition, BA + AC : AC: : BC: DC. This analogy gives DC, and BD is then found by sub
PROP. C. THEOR. If an exterior angle of a triangle be bisected by a straight line, which cuts the base produced ; the rectangle contained by the sides of the triangle, and the square of the bisecting line, are together equal to the rectangle contained by the segments of the base intercepted between its extremities and the bisecting line.
Let the exterior angle CAF of the triangle BAC be bisected by AD: the rectangle BA.AC, and the square of AD are together equal to the rectangle BD.DC.
Describe (IV.5.) the circle AEBC about the triangle ABC: produce DA to E; and join EC. Then, since (hyp.) the angles CAD, FAD are equal, their supplements CAE, DAB (I. def. 38. and ax. 3.) are also equal; and (III. 21.) B and E are equal. Therefore, (VI. 4. cor.) in the triangles BAD, EAC, as BA : AD : : EA : AC; and consequently (VI. 16.) the rectangle BA.AC is equal to EA.AD, To each add the square of AD; therefore BA.AC + AD= EA.AD + AD", or (11.3.) BA.AC + ADP = ED.DA; or (111.36.cor.) BA.AC + AD = BD.DC. Therefore, if an exterior angle, &c.
PROP. D. THEOR. IF from an angle of a triangle a perpendicular be drawn to the base ; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.
Let ABC be a triangle, AD the perpendicular from the angle A to BC; and AE a diameter of the circumscribed circle ABEC: the rectangle BA.AC is equal to the rectangle AD.AE.
Join EC. Then the right angle BDA is equal (III. 31.) to the angle ECA in a semicircle, and (III. 21.) the angle B to the angle E in the same segment; therefore (VI. 4. cor.) as BA : AD :: EA : AC; and consequently
traction. But by this proposition BA.AC=BD.DC + AD”, therefore, from the area of BA.AC, take that of BD.DC, and the square root of the remainder will be AD.
In a similar manner, from propositions A. and C., the line bisecting the exterior vertical angle may be computed : and from proposition D., in connexion with tbe 12th or 13th of the second book, the diameter of the circumscribed circle may be computed, when the sides of the triangle are given in numbers.
(VI. 16.) the rectangle BA.AC is equal to the rectangle EA.AD. ìf, therefore, from an angle of a triangle, &c.
PROP. E. THEOR.
The rectangle contained by the diagonals of a quadrilateral inscribed in a circle, is equal to both the rectangles contained by its opposite sides.
Let ABCD be a quadrilateral inscribed in a circle, and join AC, BD; the rectangle AC.BD is equal to the two rectangles AB.CD, and AD.BC.
Make the angle ABE equal to DBC, and take each of them from the whole angle ABC; then the remaining angles CBE, ABD are equal ; and (III. 21.) the angles ECB, ADB are equal. Therefore (VI. 4. cor.) in the triangles ABD, EBC, as BC: CE::BD: DA ; whence (VI. 16.) BC.DA= BD.CE. Again, in the triangles BAE, BDC, because, (const.) the angles ABE, DBC are equal, as also (III. 21.) BAE, BDC, therefore (VI. 4. cor.) as BA : AE:: BD: DC; whence (VI. 16.) BA.DC = BD.AE. Add these equal rectangles to the equals BC.DA and BD.CE: then BA.DC + BC.DA = BD.CE+ BD.AE, or (II. 1.) BA.DC + BC.DA = BD.AC. Therefore the rectangle, &c.
Cor. 1. If the sides AD, DC, and consequently (III. 28. and 27.) the arcs AD, DC, and the angles ABD, CBD be equal, the rectangle BD.AC is equal to AB.AD together with BC.AD, or (II. 1.) to the rectangle under AD, and the sum of AB and BC. Hence (VI. 16.) AB + BC: BD :: AC : AD or DC.
Cor. 2. If AC, AD, CD be all equal, the last analogy becomes AB + BC: BD:: AD: AD; whence AB + BC = BD. Hence in an equilateral triangle inscribed in a circle, a straight line drawn from the vertex to a point in the arc cut off by the base, is equal to the sum of the chords drawn from that point to the extremities of the base.
PROP. F. THEOR.
The diagonals of a quadrilateral inscribed in a circle, are proportional to the sums of the rectangles contained by the sides meeting at their extremities.
Let ABCD be a quadrilateral inscribed in a circle, and AC, BD its diagonals; AC:BD :: BA. AD + BC.CD: AB.BC+AD.DC.
If AC, BD cut one another perpendicularly in L, then, (VI. D.)