yay

Question 1

Answer 1

y=2x
< at centre twice < at Oce

Question 2

if AB is a diameter -> 90!degrees

Answer 2

< in semi circle

Question 3

90 degrees -> diameter

Answer 3

converse of < in semi circle

Question 4

x=y

Answer 4

<s in the same segment

Question 5

x+z=180

Answer 5

opp <s, cyclic quad

Question 6

x=y

Answer 6

ext < cyclic quad

Question 7

proving concyclic -> <p=<q or <s=<r

Answer 7

converse of <s in the same segment

Question 8

proving concyclic: <s+<q or <r+<p = 180

Answer 8

opp <s supp

Question 9

OP ⊥ AB -> AP=PB

Answer 9

LFC,CBC
line from centre ⊥ chord bisects chord

Question 10

AP=PB -> OP⊥AB

Answer 10

LJCTMOC,C
line joining centre to midpoint of chord ⊥ chord

Question 11

AP=PB and RP⊥AB. -> PR Passes through the centre of the circle

Answer 11

,BOCPTC
⊥ bisector of cord passes through centre

Question 12

OC⊥AE, OD⊥BF and AE = BF, then OC= OD.

Answer 12

equal chords, equidistant from centre

Question 13

OC⊥AE, OD⊥BF and OC= OD, then AB = BF

Answer 13

chords equidistant from centre are equal

Question 14

if AB=CD , then x = y

Answer 14

equal <s, equal chords

Question 15

if arcAB=arcCD, then x = y.

Answer 15

equal <s, equal arcs

Question 16

if arcAB=arcCD -> AB=CD

Answer 16

equal chords, equal arcs

Question 17

arcAB:arcCD = x:y

Answer 17

arcs prop. to <s at centre

Question 18

arcAB:arcCD = x:y

Answer 18

arcs prop to <s at Oce

Question 19

if AB is the tangent to the circle, then AB is perpendicular to OP

Answer 19

tangent ⊥ radius

Question 20

if AB is perpendicular to OP, APB is the tangent to the circle

Answer 20

converse of tangent ⊥ radius

Question 21

tangent properties

Answer 21

1) AT=BT
2) <BOT=<AOT
3) <OTB = <OTA
4) ABOT are concyclic

Question 22

if AP is a tangent to circle at P, x=y

Answer 22

< in alt segment

Question 23

if x=y, then AP is a tangent to the circle at P

Answer 23

converse of < in alt segment

Question 24

q1prove similar

Answer 24

<ABE=<CDE (<s in same segment)
<BAE = DCE (<s in same segment)
<AEB=<CED (vert. opp <s)
ABE ~ CDE (AAA)

Question 25

q2 prove similar

Question 26

q3 prove similar

Question 27

converse of base < isos triangle

Answer 27

sides opp equal angles