真二第張六第日五十月四年UNWAH KIU YATO
日十三月五年九六九一曆公年八十五國民華中 育教僑
O is the centre of the
given semicircle (touches.
AC up)
And.
LAOT
<BOT CATO BTO
Com, 4s of a Ag
育数僑華
OT bissch
'LADB' and
< ATG
JZ(b) "Give.
AB= AC BO=0L
英中會考數學(三)答案
(續)
歐陽鉈女:
M
Mathematics
C. E.E. (English)
Paper I Suggested Answer
Section e
AB,
MN is a tangent. to the semicircle.
To Prove: < BOM = «[NO
Proot
Let the semicircle touches the simtes
at P. Q. R esp.
and let BOP=2, ¿POMAL «MOQ=C 4Q0N=4, « NORTE, <ROC=f.
Join OP, 0Q,
BEHI
拨
AF LFG
E.G
Proof. Join ER
HEII FS
EF LFG CHEF-4EFG. = 40
HEF is a at E
Given
alt, és HEIFG
of which HF is the hypotenus
K is the mid-point of hypotenuse MJ. The circle, central K, alia =
= Hy passes through E
A. KEHK = K=
11, het ? be a fixed point inside a
whose centre is
circle
(Q) State the locus of the mid-points of the
chords through P
by Prove this statement in two steps as followe
(s) that every point which satisfies the conditions
hes on the
that ever
OR.
(The proot here is given in outline form) O is an ex- centre of ABC.
But
HK = EG
(Given)
[RHS)
in L. ¿EK6
10 cas
point which res on the focus
(3)
A DPB = 4 ORC
a = t
(4) From Dart (a), ari bistako «PDQ, bac On bisick 400R ̧ ¬d=e
(5) Hence, a+b+d=c+ ert
- 1 of 1800
- 90°
-". a+b
Satistir the concutions
Proof: (as Let AB, CD be any
two arbitrary, choras
which
pass thru pt.
St. MiN are the mid-points of
Chords AB, CD resp
Join DP DM. ON
<OMP=LONP (=90°).
OAP, MN
che con cualit
dince
<OM P = LONP <Ÿo*
OP
D
1.2.
< EDM is the complement of a (6) in 4. A OQN, «ON&= $p°-d (7) By part (a), ▲ ONC, = LONQ =90°-d By pt. () and (7). we find
< BOM=/CNO (= 90°-d)
KE = EG
EK = EG E6K = LEK6
EKG = «H+e
KH KE
1.-4=e
proven
base, 4; 180s a
ix z of proved
base 41, 1805
LEKG= ALM
·EGK = CEKG = P_H
alt 4. HE II FG.
And HGF= <H
LEGF=
=
- t6k + <HGF
3 x H
4E6 F = 3° «#GF.
10
< H&F÷ ÷ 4ESF
15, Given FEG HABI CO
13) Given: Quad ABCD
1,00 in
Pam concyclo
circle with dia
the lo cus
Hence
otra me ter
curce of
the required locus is a diameter Of
(by Condition
ma-pe of the chord & thru ja a circle with diameter OP
w Let k be the mid pr of arbitrary choya.
XY which satisfies the condition
Then,
~
OK LXY
« OKPI 96′′
the line jouniting the centre to the mid-jt of a chord I chord)
QNH3 LOrp=90"
k lies on the
witę AM=MU
BN-ND
OM I/BD, ONIAC AP=PD, AQ=08
To Prove = 10) AIMQ=#POQ
(b) A PoŹRBCD,
P
To Prove
To find.
·Proot: (á) vem FQ
AP=PD. 92=Q6
Given
Solution
PQ / BD
Since
OM BBD
mid-24thesem Given
(0) 4 4, CAB, CEG
EAC, EBD FAN
and GBC cure
st Limes
b. the distance
:
bm FG+ AB-
h2 the distance
buto CD AB
FE E-G
FG in terms of h, h.. and AB.
Let AB=5. FE=x. E 6=y units.
48 #EG
(Given)
CA
(1)
PQ IOM
CE
·PQ 1[OM
•proved
In A. DAB. DFE
98 // FE (Guren ›.
As on
the
same bais (PQ) and
I
DB
= OP
K
un If K' ra
an arbitrary point on the Cncu, meter OP
between the same paratbils (PQ #MO)
2. A POM= A PRO
APMQ = ~ APQ+A P Q M
APOQ= ~ APQ + APQO
of which a PQM = APQO. (proved!
Bund APMQ
quad. APOQ
In S. FAB. ECD...
DE
ME /CD
( GIPEN]
EA
Ec
ED
EA
D
EC-EA
FD-E8
EC
ED
AC
BD
EC
ED
Join Ok
OP 13
ZÁ Z OK Y
Xia mela
(b) Join PC.
- in semi Carco.
A, PAM, POM RAE
and of the
equal bases (AM=ME)
from (1) and tha
Some altitude (from vertex Pi
- APAM = AFMC
* me Chora passes thru Pe
the i from centre to a chord will besart the chord
ok' bisects X'X'
x' is the me point of Xy
* K* satisfies the Canolitrớn.
From () and tin the circle with diameter or
is the required locus
1280) Given: T is
a point
Outside circle O.
TA, TB are tangents from T
To Prove: To bisect <AOB
ATB.
Proot Join OA.OB. OT
radii of O
B
are tangents from T to 00.
- OA, OB arre
TA. TE
OAL TA
JA, OAT OBT:
OR LTB
1
<OAT - 40BT =40*
OTOT
.DA OB
COATS A08T
es CAP
CDP war on equal basss (AP = PD) and of the same altitude (from vertex ()
SCPA = adDr.
(i) Hence, AYAM = SIMC • ± of CPA=DADC
Similarly, join MB
***
As MAQ, MQB an on equal basts (AQ=QB)
und of
the
Same
altitude (from wertex M)
2. AMAQ FAMOR — of AMAB
As BAM, BCM are on equal bases (AM=MC) and of the same altitude (from vertex B)
.. AMAR = a MCB = Lot A BAC
Gi> Hence,
tia +us?
▲ MAQ ·± of = MAR = 40BA1 APMQA PAM + AMÁQ
· 4a ADC + ABHC
- ABC)
provest
COMMEN
badii
From (a), we proved that APOR - APMQ
APOQ = + ABC D
RMS,
E
+ = CA. 1 - 21
올 - 좇
X
FE
= EG
(b) From
4. DAB. DFE
DAB
<< DEE
Cor.is ABIFE
in
Commen
Z D
ODAB
ДАВ ADFE
A OFE
p(hth)
By la),
x = y
FG = I
h. th.
ap (4, the)
16, (4) Given ·
-
6 ABC with AB > AC
median
is produced to
AD is
4 D
E
3
TO PROUD
Proot
DEAD
- CAD > BAD
Join BE
( 未完轉入第六張第三頁)
No comments yet.
Private notes are available after approval.