.. gnomon AKF + fig. CK 2. AB. BC, add to each the fig. HF, which is equal to AC2, :. gnomon AKF + fig.) CK + fig. HF but gnomon AKF + fig. CK +fig. HF .. whole fig. ADEB + i. e. fig. CK 2. AB. BC+AC2, whole fig. ADEB +fig. CK, = 2 AB. BC+AC2, ABBC22 AB. BC + AC2. Wherefore, if a str. line, &c. Note.-If the parts of the line be a and b, the proposition algebraically expressed is (a + b)2 + a2 = 2a (a + b)+b2. Ör (a + b)2 +b2 = 2b (a + b) + a2. PROP. XLVII. THEOR. 12. 2 Eu. In obtuse angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle, is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle. Let ABC have the obtuse ACB. From A draw ADL BC produced: then BA2 > BC + CA' by 2 BC. CD. str. line BD is divided into 2 parts in pt. C, BC2 + CD + DA2 + 2 BC. CD; BD2 + DA2 = BA2, CD + DA2 = CA2, i. e. BA2 BC+CA2+2 BC. CD, Therefore, in obtuse angled As, &c. Prop. 39. PROP. XLVIII. THEOR. 13. 2 Eu. In every triangle, the square of the side subtending either of the acute angles, is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle. Let ABC have the acute ABC, upon BC one of the sides containing it, draw AD 1 BC; then AC2 < CB' + BA by 2 CB.BD. (Fig. 1.) Prop. 46. .. ... str. line CB is divided into 2 parts at D, CB2+ BD2+AD22.CB.BD+DC+AD, Prop. 39. i. e. Prop. 15. Prop. 47. Prop. 42. Prop. 39. CB2+ BA2 = 2 CB. BD+AC2 2ndly. When AD falls without the ABC. (Fig. 2.) D= rt. L,: LACB > rt L AB AC+ CB2+2 CB. CD, add CB2 to each, .. AB2+CB2 = AC3+2CB2+2CB. CD. BD is divided into 2 parts at C. BD.CB CB2+ CB. CD, .. 2 BD. CB= 2 CB'+2. CB. CD, .. AB2+CB2 = AC2+2 BD. CB, i. e. AC2 < CB2+ AB2 by 2 BD. CB. Lastly. When AC BC, then is BC the str. line between the perpendicular and the acute at B. (Fig. 3.) And AB AC2 + CB2, add CB to each. .. AB2+CB2 = AC2+2 CB2, i. e. AC2 < AB2 + CB2 by 2 CB. CB. Therefore in every A, &c. DEF.-A straight line is said to touch a circle when it meets the circle, and being produced does not cut it. This line is said to be a tangent to the circle at the point of contact. PROP. XLIX. PROB, 1. 3 Eu. To find the centre of a given circle. Let ABC be a given ; it is required to find its cr. In the draw the str. line AB, and bisect it i. e. greater less, which is impossible. = And no other point but F is the Cr.; which was to be found. COR.-If in a O, a str. line bisect another one at rt. s, the Cr. of ing line. is in the bisect PROP. L. THEOR. 3. 3 Eu. If a straight line drawn through the centre of Let ABC be a O, and let CD passing through the Cr. bisect AB not passing through it, in the point F; then CD shall cut AB at rt. s. |