circle

blah blah blah

1b)
<BCD=180°-y (opp. <s, cyclic quad.)

u r correct =.=
In BDC,
<CBD + <BCD + <BDC = 180° (< sum of triangle)
35° + 180°-y + <BDC =180°
<BDC=y-35°

c) ii)
if CD=CB,
<CDB=<CBD=35°
<BCD=110° (by < sum of triangle)
180°-y=110°
y=70°

if BD=BC,
y-35°=180°-y (base. <s, isos. triangle)
y=107.5°

2a)
BCDE is a concyclic quadrilateral obviously.
·.· B, C, D, E lie on the circumference of the circle.
<CDE+<CBE=180° (opp. <s, cyclic quad.)
<CBE=85°
·.·BE=CD
.·.<CBE = <BCD = 85° (base. <s, isos. triangle)
·.·<BCD + <CDE = 180° (int. <s, supp.)
.·.BC//ED

b) <CDE=<ABE=95° (ext. <, cyclic quad.)
In triangle ABE,
<BAE=45° (by < sum of triangle)

c)
·.·ABC is a st line & BC//ED.
.·.AB//ED
(If you think this is too short, then use
<BCD+<BED=180° (opp. <s, cyclic quad.)
<BED=95°
·.·<ABE=<BED=95°
.·.AB//ED (alt. <s, supp.) =.=)
Join BD.
<AEB=<EBD=40°(alt. <s, eq.)
In triangle BED,
<BDE=45° (by < sum of triangle)

2012-05-04 04:42:00 補充：
-.-
2c)
.·.AB//ED (alt.



2012-05-04 04:43:31 補充：
-.-
2c)
.·.AB//ED (alt. angle s, ** eq.)
Join BD.


2012-05-04 04:45:17 補充：
............


2012-05-04 04:50:06 補充：
2c尾三尾五兩行自己轉返岩 = =

2012-05-04 18:22:39 補充：
BE=CD
angle CBE = angle BCD (base. angles, isos. triangle)
定義黎

如果唔係用 (base angles, isos. triangle) 既話
佢比BE=CD做咩

2012-05-04 18:28:41 補充：
你覺得唔岩既咪用alt. angles, eq.黎prove BC//ED (join BD)
一樣得

第1題thx,比佢個shape FAKE左==
第2題,.·.


2012-05-04 13:31:43 補充：
angle CBE = angle BCD = 85° (base.

2012-05-04 13:32:28 補充：
angles,isos. triangle)
點解會咁?