20(a)Angle DCE=90(BC perpendicular to CD)
Angle DAE=90(tangent⊥radius)
Angle DCE+Angle DAE=180
AECD are concyclic(opposite angle supplementary)
20(b)
angle BAM=angle ACB(angle in alt. segment)
Since ADCE is concyclic,angle ADE=angle ACB(angle in same segment)
21(a)Angle ADB=90(angle in semi circle)
x+Angle ADB+Angle CBE=180(opposite angle,cyclic quadrilateral)
Angle CBE=90-x
21(b)x+90-x+Angle CEB=180(angle sum of triangle)
Angle CEB=90
CE perpendicular to AB
21(c)
angle ADF + angle FEA = 180
ADFE is concyclic(opposite angles supplementary)
23(a).Angle SOP = 2 Angle SQX (angle at centre,twice angle at circumference)
Angle ROQ=Angle SOP (Equal arcs, equal angles)
Angle RSQ = (1/2)x Angle ROQ = Angle SQX(angle at centre,twice angle at circumference)
Angle SXQ= 180 - 2 Angle SQX (angle sum of triangle)
Angle QXR = 2 Angle SQX(adjacent angle on straight line)
(b)Angle QXR = Angle SXP (vertically opposite angle)
Angle SOP = Angle SXP (proved in part a)
PSOX is concyclic(converse of angle in same segment)
22(a)
Angle APQ=180-a(opposite angle,cyclic quadrilateral)
Angle APQ+c=180(adjacent angle on straight line)
180-a+c=180
a=c
(b)Angle XYQ=180-b(adjacent angle on straight line)
b=c(angle in same segment)
c=a(proved in part a)
Angle XYQ + Angle XBQ = 180-a+a=180
BQYX are concyclic(opposite angle supplementary)
Edited and finished
[ 本文被hohoho2在2010-10-28 18:02重新編輯 ]
