賞一第張五常
WAH KIU YAT PO
fa
育教僑華
一葉中文中學會考試題預習專欄
英文科 (廿三)
·英。
TYRSKY
英文中學會考試題預習專欄
***
數學科 (十三) ·歐陽鐇文·
MAT:IRMATICS (23)
LESSUN 23 CIRCLASS
EXAMPLAS 1: ( en cheras er a elwale)
X, I are the mid-points of the cherds
AB, CD of a circle, centre O; IN, YM
are the perpendiculars fram X, I to
CD, AB resp.,If X cuts at F,
prove that OP and XT bisect sach other.
PROOF: Join OX, OY.
Y is the mid-point of the sherd ABN
OLING
* OXLAB.
- 1 AB (given)
0X//IM
Similarly. beth OF, IN are perp. Le CD.T: OY//XF.
· OX // FT & ÖY {/ IP-
By definition, PYOX is therefëre a parallelogram. Hence, OP and XY bissct each other,
EXAMPLE 2: angle properties of a circle)]
The bisectors of the angles ABC,
ACB of ABC intersect at I and
cut AO, AB at Y, 2 resp.; the circles BIZ, CIY meet again at E. Prove that IXZ+/BIC=2 rt./s. ROOF: With the netāțians as shown..
Join II.
/TXY = c. is in same deg. } ZIXZ = b.
Since, .-,and b, b2 (given)
[YXZ = [XXY → [IXZ - “.+ b,
From ABC, by / sum of a
BIC + b2+ c2-2 rt.₤s
· / BIC + /1X2 = ? rt./s
SXAMPLE 31 ( concyclic points)
ABCD is a parallelogram.
O is a pt. inside ABCD auch that,
LAGB + LGOD = 2 rt.Lo
Prove that 40BG – 4 0DC.
2007: Draw AP DO, BP # CO.
Since the three sides of AAB?
are parallel to the sides of AÇDO
• ABP ÷ + D60
But, ABCD
AABP RADCO
LAPB #DOC
app. #idesgran.
ASA
corr. fa oỈ KHÔNG
LAPB+LAOB/DOC + ₤108 = 2 rt./#
A, P, B, O are concyclic
OC PP constr. & preved OP # BC α= b
Join OP,
is is same segment
-proved db. 1,9. /OBC = 1000.
in whieb, and, & memb
werk24% (chords & arss)"
ABG inscribed in ́s circle as enéwn, Dis the mid-point of minor are BC. O is a point en DA such that DO DG. Prove that 0 is the incentre of aABC. FROOF: Join 00.
miner ard EDwainer are DC (givan)
In the eare circle, equal ares subtend
equal angles (at y
1.5. OA is the bisectar of ¿BAG.
It is required to prove that OC is the bisector of LC.
In AOAG,
In
OCD,
ext./ of a
base 8, is00.
But ¢ ̧—à ̧~8, (is in saus segment)
** 0,= ©2
1.0, oc la the bisector of [BCA. Fram (ə) ado (b), we conclude that,
Dis the in-centre of ABV.
EXAMPLE
ABC 13 an equilateral triangle inscribed in a circle. P is any point on minor arc
BC. Prove that: PA~ FB + PC
400F: Produce BF to D,such that PD = PG
In ACPD, PC= PD
constr.
[CPD-BAC60 ext.[of cyclic PCD is equilateral. quad.
Consider 4APC, BCD:
LACP BCD (= 60 +, ∞ ).
AC=BC
GP CD
A ACP
BCD
(DC-DC)..
-(b)
↑ KALEVSTA "To me ̃previous"exeróisa;
}*] B00GUGS TO: GOUIN BOT.Ko: to, plas]like the others. He had to,
| WAL TOWARD Ina TAROP +
DATA
rance that he was going;to:vbitowäed-made his heart:filled)
with despair
to.1rno immense unwhite-vasbed:fence made bio it on the gox
• di scouTALSU.
*A get'water from"the"town"pump›
Doma |
.. Bo. be used to
I.
DO. 31 was not
Far
Away. It"vs"only"a" hundred "and" fifty yard.
away. But i did not retum, free the Dund at once. Na wang
to fool
around with the boys st the puan.. That was why be.
had to take as nour to. do the NOTAI
I. Too aaked Jim to leX DID O prat STom: the: pung}
↓
ne was tempted by Tom's marviT.
To, because he ese stopped by Aunt Polly.
1. BennUL DA ADA ВАЛ СОда in the right time.toletop: Top true,
running away from his work_
BOO
k. The thought that the free boye would
worm mini 020.2158 150.ĺ
J
ПЕТІВК FOE
i. Dals we having no dope.
dipped
Тормов
innsreed.
the very;top
opera 1 On A g
tririing
CRIAD
immen 81ty
WORK
emaxxi
Long 1 no.
TOO1232 abou
- great amoun
BIỆT - truk
VOITO-a §b
energetica V
umphant
----} Joustne y w;
二期
日四月四年七六九一屉公年六十五——
En room," or even coletanos-elastinTAAN malenTANA KAN (well quite worthily." We can saamaikatġur|Close the genuine bearçon va rewa»»« đo yords tali us.) There¡ie us suku.rineco, no restraint, on either side. The thread of our intimacy bas been snapped. {The leave-taking is an ideal (ma.my co7, (then leave the leave-taking at that? Always,įdoparting friends Implora us not to bother to comA the railway station nexe] morning." Always, we zre deaf to theee entraatioë, knowing thea] to be not quite sincere. The departing friends would think is very odd of us Yo took them at their word.] Besides, they, really] do want to see us again. and thas wish is heartily reciprocated;
to
we duly turn up. And then, oh then,' chat a gulf! yawna) Vel
*
stretch our arms vainly across it. ve hava utterly lost touch
We have nothing at all to say. Ve gaza s1 each other as animale gaze at bunad beinga. Vej'cake conversation'----} end eüch
-Jend
conversation! We know that these are the friends from whom we
parted overnight. They know that we have not altered. Yet,|| the surface, everything is different, and the tension is such
that we only long for the guard to blow his whistle, and put. Any
end to the farsa..
·Hintay
feat
lamentably
in ratio to
1
restraint -
DL:49 flange WW3
entreaties
11.
reciprocatad
f
farcel
MATH
2.Translate the following incathellahi
7
vinh thể önen kandi
tq # #
cover witda mixture?
full or energy, mocessful.
Exerc186 23
Trungista on
1. Translate the following into Chineser
2 GLID po's good as zvito go 11 wall seemo"to"më ons on
the most darriault zaings in the world.and probably seems sol
to you too.
To see a friend.
enougn. Bus ve 379 nov
It is only when a trio
wari
De Bosent for a 12
EDAIWAN Ętations The Gearez,
Frailvar atation were easy
? to perfori tbit anali feat.
ob.aux.ongien Journey O
that we turn up at the
lema and the zonama %D8
likely sokeace, the sarller do we tum up, and the core lament¬
Die ad ve rail. Our failure ta in orast Patia to the terious;
to the depth of our feeling.
nose of tas occasion,
1
静らみ利に好
性嗜酒「家貧不能常得親會知其以此置
ORK HAMMMEL TAN
ABNIKIO
BRARK
generally mosh;am
(nonour "and"riches:
surrounding walls
Į DASİFƏTƏ ̧ANË, KOUSENĮ Corsźre-save(iyatr
8
AP BD,
$ino, 80=BE+ PD =BP÷+ PC
sides of squí. a PCD.
SAS
HOTE; We may také s polat E on AP such that PE=FB.Hence, prove that, BPC AABE LA BPB de equilateral ). then PC AE.
EXAMPLE 6: (tangents)
As shown in fig. 1,8 4re
the centres of two circles
touching the parallel lines
GQH & KPF; the common
tangents PQ touches each
circle as shown. Prove that
(a) AFBQ 16 a rectangle;
(b) PQA8.
$ROOP: Since PK,PQ are tangenta from P to circle A.
AP bisects EPQ.
Similarly. BQ bisecte PQ8.
But
EPQ/PQH
**paq ̈·
AP/78Q
AQ//BP
& AQ//BP.
alt./g.CH//BP)
alt-10 p=9.
Q
Similarly. AQ,BP are the bisectors of the alt.La GQP,QPF,
APBQ is, then, a //gram.
44, 80 ere bisectors of two adj./s on at. Line CQH.
And,
ZACB of 2 rt./a 1 rt./
•APPO de therefore a rectangle, AR=PQ (diaz, of a rect.)
SEAMPLE 7+ (alternate segment)
"ABCD 18 a minor are of a cirozė sucoj
that AB BC
when AB and DC anst "produced, at P, and DB is produced -to pest the tangent if at T.
Prove that TP-TA:
AB BG
stand on equal chorda era equa.
d,- da
At Le a tangent et å
is in alt. sagmen
PROOF
..
A. T, P, D are concyclic
¿TPA de
{ TPA (4,)'
XAMPER 81 (contact of circles)
Vin ARC, AB=p, AC=q, {BAC = 90° and pg:0 is the mid-polat of BC. Gizeleaseze drawn with AB &
AC as diameters.
Prove that two circles can powe drawn with as centre to touch each of these circfes, and,
"/a Lo same segment?
eides oppɔ, equal je
find their radii in terms of p & ą. PROOF: Let E, F be the mid-points of 48, au resp.
Join OR and produce it to cut OE at H, K. Join OF and produce it to cut OF at K, N.
NOTE
OH=QE+J
where OE»‡AG (old-point theorem)
EH-EA
AB
OR( AB+ AC) - -
Similarly, O=OF+ FM
whers OP † AB and FM÷FA AC
OM (AB+ AC)
from (a) and (b), we have OH÷Œ,
Hence, with centre 0, radius OH −£{p+q} We can one such circle,
Secondly," consider now, 08 and 0X.
-
OK EK-XO -- (AB AC) = ±(p −9} ON=0F-HF =÷(AB AC) (p) Hease, with centre 0, radius of fipin), HE ĐI (another circle to touch each of the given öirales,
FO
As shown in figure:
|(9)-
(1) ABC. Is said to be giroumscribed
about e 0.
:
(11) Quad, LMFP is said to be inseribed
in 0.0.
(111) ia dala to be th
circumscribed about IMEP) or, inscribed in a ABC.
0. D ́are în-cushe " ex-owne of DABC
ADDIE MET. 16"
DOA, FAE are the int, 4
angh - brector of 44.
DOA FAE
20h is an
ADEF
Similarly § EDB, rocTM
OF LA DEF.
T'a cu mest, at a
the artho conte DIES
ABDC = ABK}(SAS)
db,
A CDLBK.
Similarly, BFLECA
• KH BF CD an athhudes of akse. 12. AH, BF, CD are con Current
Cat the orto centre of a KBC),
Sa Ket the perp biztatus of 16, BC, AC
meet at o ( sircumcentre of a BBC)' 14 M AÐ, BY wet at P AC, AD, CD MEET NË Q
Then, Pro both lie en de purp bisector of AB, Q TO LOA # on Mo
perp. bisector of AC etc.
P. & bor, cäincuda with O.
NOTE. This pros. in very easy by applying the properties)
of chords in a carele.
The perp bisects of a cand passes through
the
DEMO.
Hence, all ML Con current at the cente
centre of the circle bisi ctos
AZLHUISE 23
are
c) Proof: Join BD.
OD ADC QB 16C 'Hence, Q is the
SABD
DQ produced & R
Bd produced & BD.
orthocentre of
AR 1 BD
a) A triangle ABC is inscribed in a cirole, and the bisectors.
of the Za meet the
at X, Y, Z. Show that the Is of XXL are resp., 90-£; 90 ; 90-£.
(2) Two circles meet at A. and B as show, 40, AD are the diameter,
of each circle. Prove that:C, B, D are collinear.
Two circles meet at A, B. CD is a common tangent to LWO circles. Prove that CAD / CBD-180.-
ABC is inscribed in a circle. O is the orthocentre of ARBU, The thres altitudes AD, BE, CF are produced to meet the circle again at G, H, X resp. Prove that O is also the In-centre of GHK
•
(5) Two equal chords AB, CD meet, when produced, at u.
Prove that BGDG
ABCD is a cyclic quad. If M, N, P, Q are the centres or rom circles inscribed in As ABC, ABD, ACD, BCD. Prove that NFQ [\is a rectangle.Me
The diagonals of cyclic quad, ABCD out each other, at re.fs at P. Prove that the prependicular from P to BC. bisects, when produced, ¿D. (Brachmegupts Theores.)
#) AOB, COD are two prependicular diameters of a circle, Tw chorde CP, Ca out AB at H, I, Prove that H, K, Q, P.... ars coneyclic.
Figure for questions A