A vector is defines from A to B[br]A vector is a line with direction, it requires a starting point and an ending point.
Figure 2.1 shows Vector AB, Vector CD, and Vector EF.
Using Figure 2.1, manipulate points A through F. As you move each point, observe how vectors AB, CD, and EF change. Record your observations.
EF always the same length as AB (mark 1)[br]EF always parallel to AB (mark 1)[br]EF always opposite in direction to AB (mark 1)[br][br]Vector EF = - AB (mark 2)
A circle can be drawn with any center and any radius.[br]The circumference is the total distance around the edge of a circle. It is the perimeter of a circle.[br]The radius is the distance from the center of the circle to the circumference.[br]A diameter is a straight line segment that passes through the center of a circle with both endpoints on its circumference.
Circle AB is defined by the radius AB with center at A[br]Circle CD is defined by the vector CD with center at C[br]Circle F is defined by the center point F
Analyze the arrangement of circles as you manipulate the points. [br]Move points A through F and observe how the circles change.
Pull on point B and both AB and F change size. (1 mark)[br]Pull on point D and both CD and F change size (1 mark)[br][br]The diameter of circle F = 2AB +2CD (2 marks)
A tangent to a circle is a straight line that touches the circle at exactly one point, known as the point of tangency. The tangent of a circle is perpendicular to the radius at that point.
Circle AB is centered at A with radius AB. Circle BA is centered at B with radius BA. Circle CB is centered at C with radius CB. Circles AB, BA, and CB are all equal in size, they are said to be congruent.[br][br]Line DE passes through B and forms a tangent to circle AB and circle CB at point B.[br][br]Looking at angles ABD, DBE, EBC, and CBA, we can see they equally share the 2D plane. [br][br]By Euclid's definition, Line AC passing through B is perpendicular to Line DE passing through at point B.
Cut a line in half with a 90 degree angle
AB is the radius of both circle AB (center A) and circle BA (center B). The two circles intersect at points C and D.[br][br]Line AC, AD, BC, and BD are all equal in length to AB (the radius of both circles).[br][br]Based on similar triangles:[br]Triangle ABC is congruent to triangle ADB by SSS.[br][br]AD = AC[br]DB = CB[br]AB = AB[br][br]Therefore, angle DAB = angle CAB. Since angle DAB + angle CAB = angle DAC,[br] then angle DAC = 2 * angle DAB.[br][br]Likewise, angle CBA = angle DBA. Since angle CBA + angle DBA = angle CBD,[br] then angle CBD = 2 * angle CBA.[br][br]We can say that line CD cuts line AB in half and E is the midpoint of AB.
Complete the proof to show that CD is perpendicular to AB.[br][br](Hint: Use the congruent triangles AEC and AED, or BEC and BED, to show that angle AEC = angle AED = 90 degrees.)
(1 mark) States that E is the midpoint of AB.[br](1 mark) States that triangle BEC is congruent to triangle BED by SSS, using BC = BD, CE = DE, and BE = BE.[br](1 mark) Concludes that angle BEC = angle BED.[br](1 mark) States that angle BEC + angle BED = 180 degrees, so each is 90 degrees. Therefore, CD is perpendicular to AB.
Circle A has center A, with points B and C on its circumference.[br][br]Circle B and circle C intersect at point A (the center of circle A) and point D.[br][br]Circle B (center B) and circle C (center C) each have the same radius as circle A.[br][br]Line AD cuts chord CB at its midpoint.
Prove that AC, CD, DB, and BA are all equal in length.
(1 mark) States that AC = AB (both are radii of circle A).[br](1 mark) States that AB = BD (both are radii of circle B, since circle B has the same radius as circle A).[br](1 mark) States that BD = CD (both are radii of circle C, since circle C has the same radius as circle A).
Prove that AD is the perpendicular bisector of chord CB.
(1 mark) States that AC = AB (radii of circle A) and CD = BD (radii of equal circles), so points A and D are both equidistant from C and B.[br][br](1 mark) States that any point equidistant from C and B lies on the perpendicular bisector of CB.[br][br](1 mark) Therefore, line AD is the perpendicular bisector of CB.[br][br](1 mark) Concludes that AD cuts CB at its midpoint and is perpendicular to CB.