# Books 3-9

The University Press, 1908
2 КсйфйкЭт
Пй бойплпгЮуейт ден ерблзиеэпнфбй, бллЬ з Google елЭгчей кбй кбфбсгеЯ шехдЭт ресйечьменп ьфбн фп енфпрЯжей

### Фй лЭне пй чсЮуфет -Уэнфбоз ксйфйкЮт

Ден енфпрЯубме ксйфйкЭт уфйт ухнЮиейт фпрпиеуЯет.

### Ресйечьменб

 PAGE 42 Book III 78 DEFINITIONS 112
 191 302 BOOK VIII 345 BOOK IX 384

### ДзмпцйлЮ брпурЬумбфб

УелЯдб 34 - EQUAL straight lines in a circle are equally distant from the centre ; and those which are equally distant from the centre, are equal to one another.
УелЯдб 65 - If a straight line touch a circle, and from the point of contact a chord be drawn, the angles which this chord makes with the tangent are equal to the angles in the alternate segments.
УелЯдб 37 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of...
УелЯдб 234 - Prove that similar triangles are to one another in the duplicate ratio of their homologous sides.
УелЯдб 64 - From this it is manifest that if one angle of a triangle be equal to the other two it is a right angle, because the angle adjacent to it is equal to the same two ; (i.
УелЯдб 209 - IN a right-angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle, and to one another.
УелЯдб 90 - EF at right angles (9. 1.) to AB, AC ; DF, EF produced meet one another : for, if they do not meet, they are parallel, wherefore AB, AC, which are at right angles to them, are parallel ; which is absurd : let them meet in F, and join FA ; also, if the point F be not in BC, join BF, CF : then, because AD is equal to DB, and DF common, and at right angles to AB, the base AF is equal (4.
УелЯдб 80 - In a given circle to place a straight line, equal to a given straight line which is not greater than the diameter of the circle. Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle.
УелЯдб 212 - ... be parallel to the remaining side of the triangle. Let DE be drawn parallel to BC, one of the sides of the triangle ABC : BD is to DA, as CE to EA. Join BE, CD ; Then the triangle BDE is equal to the triangle CDE*, * «.i.
УелЯдб 95 - In the same manner, it may be demonstrated that the straight lines EC, ED, are each of them equal to EA or EB ; therefore the four straight lines EA, EB, EC, ED, are equal to one another ; and the circle described from the centre E, at the distance of one of them, shall pass through the extremities of the other three, and be described about the square ABCD.