2. The converse of the twenty-second proposition is also true; namely, If a quadrilateral has its opposite angles together equal to two right angles, a circle may be described about it. PROPOSITION XXIII. THEOREM.-If two segments of circles (ACB and ADB) are upon the same straight line (AB) and upon the same side of it, they cannot be similar without coinciding with one another. DEMONSTRATION. For, if it be possible, let (a) III. 10. the segments ACB and ADB be similar without coinciding. Then because the circumferences of the circles ACB and ADB cut one another in the two points A and B, they cannot cut in any other point (a); one of the segments must therefore fall within the other; then in the interior segment ACB take any point C, join BC and produce it to the exterior segment in D, and join CA and DA. Then because the segments ACB and ADB are similar, the angle ACB is equal to the angle ADB (6), the external to the internal, which is impossible (c). Therefore the segments ACB and ADB cannot be similar without coinciding with one another. PROPOSITION XXIV. THEOREM.-If two segments of circles (AEB and CFD) are similar and upon equal straight lines (AB and CD), they are equal to one another and have equal arcs. DEMONSTRATION. For if the segment AEB be applied to the segment CFD, so that the point A be on C, and the straight line AB upon CD, the point B shall coincide with the point D, because AB is equal to CD; A E (a) III. 23. therefore the straight line AB coinciding with CD, the segment AEB must coincide with segment CFD (a), and therefore is equal to it (b); and the arc AEB must coincide with the arc CFD, and be equal to it. SCHOLIA. 1. This proposition is proved by the method of superposition. The equality of the arcs is not stated by Euclid. 2. Since if the circumferences of two circles coincide in more than two points, they must coincide in every part, it follows that similar segments on equal chords are parts of equal circles. PROPOSITION XXV. PROBLEM.-A segment of a circle (ABC) being given, to describe the circle of which it is the segment. the chord AB perpendicularly, it passes through the center (c), and because the line FG bisects the chord BC perpendicularly, it also passes through the center (c); therefore the center must be in G, the intersection of EG and FG. SCHOLIUM. The above demonstration is substituted for the tedious one given by Euclid, who divides this proposition into three cases, namely, when the segment is less than, equal to, or greater than a semicircle. PROPOSITION XXVI. THEOREM.-If equal angles (BAC and EDF, or BGC and EHF) are in equal circles (ABC and DEF), or in the same circle, they shall stand upon equal parts of the circumference, whether they be at the center or the circumference. CONSTRUCTION. Bisect the equal angles BGC and EHF by the lines GK and HL (a), cutting the circumferences in K and L; join KC, LF, AK, and DL. DEMONSTRATION. Because the circles ABC and DEF are equal, their radii are (a) I. 9. equal; therefore in the triangles GKC and HLF the two sides GK and GC are respectively equal to the two HL and HF, and the angle KGC to the angle LHF (6); therefore the bases are equal (c), KC to LF. And because the angle KAC is equal to LDF (6), the segment KBAC is similar to the segment LEDF (d); and they are upon equal straight lines KC and LF, therefore the segment KBAC is equal to the segment LEDF (e). Then taking the equal segments KBAC and LEDF from the equal circles ABC and DEF, the remaining segments and their arcs KC and LF are equal (f); and in like manner it may be shown that the arcs BK and EL are equal; therefore the whole arc BKC is equal to the arc ELF. SCHOLIUM, In the enunciation of this proposition and some of the fol lowing, we have inserted the words "or the saine circle," because that which obtains in equal circles must necessarily do so in the same circle. The demonstration as given above differs from Euclid's in the given angles being bisected. This is done to render the proof applicable when the given angles at the center are equal to or greater than two right angles, in which case Euclid's proof could not be applied. COROLLARY 1. THEOREM. If two chords (AB and DC) of a circle are parallel, they intercept equal arcs. Join AD. Then because AD meets the parallels AB and CD, the alternate angles BAD and ADC are equal (a), and therefore the arcs AC and BD are equal (b). A I. 29 B COROLLARY 2. THEOREM. If two chords of a circle (AB and CD) meet one another, the angle BED formed by them is equal to the angle (FCD) terminated at the circumference by the sum or difference of the arcs (AC and BD) which they intercept, according as the point in which they meet is within or without the circle. Draw CF parallel to AB (a). Then when the point of intersection is within the circle, because the straight line CD meets the two parallels CF and AB, it forms the angle A DEB equal to the internal and opposite angle DCF (b). But the arc FB is equal to AC (c), therefore FD is equal to the sum of the arcs AC and DB, and the angle BED is B D (a) I. 31. (c) III. 26, cor. 1 equal to the angle FCD, terminated at the circumference by FD, the sum of the arcs. Again, when the point of intersection is without the circle, because the straight line ED meets the two parallels EB and CF, it forms the angle DCF equal to the internal and opposite angle DEB (b). But the arc BF is equal to AC (c), therefore FD is equal to the difference of the arcs BD and AC, and the angle BED is equal to the angle FCD, terminated at the circumference by FD, the difference of the arcs. COROLLARY 3. THEOREM. If chords intersect at the same angle, within a circle, the sums of the arcs which they respectively intercept are equal; and if they intersect without the circle, the differences are equal; but if one pair intersect within, and the other without the circle, the sum of the one pair of arcs is equal to the difference of the other. COROLLARY 4. THEOREM. If two chords intersect within a circle at right angles, the sums of the opposite arcs intercepted are equal. COROLLARY 5. THEOREM. If a chord of a circle (AB) be produced till the produced part (BC) is equal to the radius, and if a line be drawn from its extremity through the center of the circle to meet the concave circumference, the concave portion of the circumference intercepted (AE) is equal to three times the convex (BD). Draw EG parallel to AB (a), and join FB and FG. Because in the triangle BFC the side BF is equal to BC, the angle BFC is equal to C (b); and because EC meets the two parallels AC and EG, the alternate angles E and C are equal (c); therefore the angle BFC is equal to E. But the angle CFG at the center is double of the angle E at the circumference (d); therefore the angle CFG is double of BFC. To each add BFC, and the angle BFG is equal to three times the angle BFC; therefore the arc BG is three times the arc BD. But the arc AE is equal to BG (e); therefore the arc AE is three times the arc BD. E A (a) I. 31. B D G (e) III. 26, cor. 1. PROPOSITION XXVII. THEOREM.-If angles (BGC and EHF, or A and D) stand upon equal parts of the circumferences of equal circles, or of the same circle, they are equal to one another, whether they be at the center or the circumference. to EF (6); but BC is also equal to EF (c); therefore BK is equal to BC, the part equal to the whole, which is absurd; therefore neither of the angles BGC and EHF is greater than the other, but they are equal; and the angles A and D being respectively halves of the equals BGC and EHF, are themselves equal. III. 26. Hypoth. SCHOLIUM. This proposition is the converse of the twenty-sixth. PROPOSITION XXVIII. THEOREM.-If in equal circles, or the same circle, straight lines (BC and EF) are equal, they cut off equal parts of the circumferences, the greater equal to the greater (BAC to EDF), and the less to the less (BGC to EHF). CONSTRUCTION. Find the centers K and L of the circles ABC and DEF (a), and join KB, KC, LE, and LF. DEMONSTRATION. Because the circles ABC and DEF are equal, their radii are equal (6); therefore in the triangles KBC and ELF the sides BK and KC are equal to EL and LF, and the base BC A III. 1. III. 26. to EF (c), and the angle K is therefore equal to L (d). But equal angles stand upon equal parts of the circumference (e); therefore the arc BGC is equal to EHF; and because the whole circumference ABGC is equal to the whole circumference DEHF, therefore the remaining arc BAC is equal to EDF. |