急! Properties of Circles 12q16

第8條要prove congruent triangle先
PN=PN (common side)
AN=BN (given)
<PNA=<PNB=90˚ (line joining centre to mid-pt of chord, perpendicular chord)
咁∆PNA(congruent)∆PNB
最後
a, PA=PB( corr. sides, congruent ∆)
b, <PAN=<PBN (corr. angles, congruent ∆)

9. 因為K係個orthocentre
BE, CD就係個Height (Altitude)
咁<BEC=<CDB=90˚
咁佢地就會係concyclic
因為converse of <s in the same segment

10
a,
<ABX =<DCX (<s in the same segment)
<BAX=<CDX (<s in the same segment)
<AXB=<DXC (vert. opp. <s)
咁∆ AXB~∆DXC
b,
Construct a line XY s.t. XY//AB//DC
<BAX=<CDX (proved)
<AXD
=<BAX + <CDX (alt. <s // lines)
= 2<BAX
Ａ－－
　＼
Ｙ－Ｘ
　／
Ｄ－－
之後就用// LINES

11.
a,
180˚ - <CBP =<ABC
Reflex <AOC = 2(<ABC) [<s at centre, twice <s at circumference]= 360 ˚ - 2<CBP
<AOC + Reflex <AOC=360˚ (<s at a pt.)
<AOC = 360 ˚ -(360 ˚ - 2<CBP ) = 2<CBP
b,
<CBP=67 ˚
<AOC=2(67˚ )=134˚
<AOC+<OAB=180˚ (int.<s OC//AB)
<OAB=46˚

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