## Books 3-9 |

### Inni boken

Side 208

... since the triangles

... since the triangles

**have two angles of the one equal to two angles of the other**, respectively , they are equiangular to one another . We have therefore only to consider the case in which the angles included by the proportional ...### Hva folk mener - Skriv en omtale

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### Vanlige uttrykk og setninger

ABCD according added alternately angle BAC antecedent applied base Book called centre circle circle ABC circumference common measure compounded consequent construction contained continued proportion corresponding course cube definition described diameter divided double drawn duplicate equal equiangular equimultiples Euclid excess extremes fact fall figure follows four fourth further geometrical given gives greater greatest Hence hypothesis impossible joined less magnitudes mean proportional measures meet method multiple multitude namely necessary parallel parallelogram pass plane possible prime prime number proof Prop proposition proved ratio reason rectangle rectilineal reference remaining respectively right angles segment sides similar Similarly Simson square straight line suppose taken taking Theon third touch triangle triangle ABC unit VIII whole

### Populære avsnitt

Side 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.

Side 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.

Side 37 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of...

Side 234 - Prove that similar triangles are to one another in the duplicate ratio of their homologous sides.

Side 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.

Side 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.

Side 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.

Side 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.

Side 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.

Side 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.