yay Flashcards
y=2x
< at centre twice < at Oce
if AB is a diameter -> 90!degrees
< in semi circle
90 degrees -> diameter
converse of < in semi circle
x=y
<s in the same segment
x+z=180
opp <s, cyclic quad
x=y
ext < cyclic quad
proving concyclic -> <p=<q or <s=<r
converse of <s in the same segment
proving concyclic: <s+<q or <r+<p = 180
opp <s supp
OP ⊥ AB -> AP=PB
LFC,CBC
line from centre ⊥ chord bisects chord
AP=PB -> OP⊥AB
LJCTMOC,C
line joining centre to midpoint of chord ⊥ chord
AP=PB and RP⊥AB. -> PR Passes through the centre of the circle
,BOCPTC
⊥ bisector of cord passes through centre
OC⊥AE, OD⊥BF and AE = BF, then OC= OD.
equal chords, equidistant from centre
OC⊥AE, OD⊥BF and OC= OD, then AB = BF
chords equidistant from centre are equal
if AB=CD , then x = y
equal <s, equal chords
if arcAB=arcCD, then x = y.
equal <s, equal arcs
if arcAB=arcCD -> AB=CD
equal chords, equal arcs
arcAB:arcCD = x:y
arcs prop. to <s at centre
arcAB:arcCD = x:y
arcs prop to <s at Oce
if AB is the tangent to the circle, then AB is perpendicular to OP
tangent ⊥ radius
if AB is perpendicular to OP, APB is the tangent to the circle
converse of tangent ⊥ radius
tangent properties
1) AT=BT
2) <BOT=<AOT
3) <OTB = <OTA
4) ABOT are concyclic
if AP is a tangent to circle at P, x=y
< in alt segment
if x=y, then AP is a tangent to the circle at P
converse of < in alt segment
q1prove similar
<ABE=<CDE (<s in same segment)
<BAE = DCE (<s in same segment)
<AEB=<CED (vert. opp <s)
ABE ~ CDE (AAA)
q2 prove similar
<ADE = <ACB (ext. <, cyclic quad)
<AED = <ABC (ext <, cyclic quad)
<DAE = <BAC (common angle)
ABC ~ ADE (AAA)
q3 prove similar
<BAD = <CBD (< in alt segment)
<CDB = <ADB (common angle)
<BCD = <ABD (< sum of 🔼)
BCD ~ ABD
converse of base < isos triangle
sides opp equal angles
