C1
*=*B**
自四初月四年申茂夏
WAH KIU YAT PO
CITY HALL
118 英文中學會考
題預習専欄
PA, PB
dit drangeats from p to circle ADEC, PA = PB
tength of dangeor,
PA * PD
數學科 (廿七)
歐陽鎝女
1.2. ᄃ
AC- BD➡ BC · DA
07
BD
MATHEMATICS (37)
LESSON 27: SIMILARITY
IMPORTANT THEOREMS,
A, Thegems on the ratios of a ▲ formed by a ♬ line 1 cis If a si. line is drawn parallel to one side of a
A
then it divides the other two sides, produced if necess ary.
proportionally. <z> If a si. live is drawn to divide too sidhs of a a in the same ratio, both internally or wash externally, then the st, lime is parallel to the think side.
** If AB and CD are bansversais of a set of parallet
linas, then the intercepts made by the parallel lines on
are in the same ratio.........
AB and on CD
**› As shown in fig. in which XY#BC,
thentic 一般
AB
(ii) - AC
AX
AC
BX CY
Cand the converse)
UVJ ABA 45-
AX
AB
Y
A X
8, Themens on the ratios famed by ongle- bištefanı
<i» The internal ̈angle- bisector of a e divides (infernally)
the opposite side in the ratio of the other sides of the triangle. ( And also the converse). sa› the external angle- bisecter of a
<1>
a divides (externally) the opposite side in the ratio of the other sides o the ingle. And, the converse iš also hue)
angas, on similarity of a:
angular dis Care similar >
<2> Three sides ( of a, are cowrespondingly) proportional.
Chhen, the es are similar >
<a> ratio of 2 sides, inc. « ( then, the As are simha 1 3. car if two a, are of equal attitudes (or basés), then the
ratio of their areas is equal to the antio of their bases
Cov altitudes ).
<5> The ratio of perimeters of two similar as is equal
to the ratio of two corresponding sides
(<6> The ratio of areas of two similar as is equal to the square of the ratio of two corresponding ste Thegems on similarity et Circle :
Intersecting chords.
tangent property
st. lines AB. CE an divided ( bath internally on
externally ? ak-X such that
XA-X6=1IC-
XD,
hen the points A, B, C, D am concyclů.
AB is divided externally at x, and if c is
not on AB, such that 24 X8 = ac* tea, the circle Agc touches. 2C) à c
Trapezium ABCD with Haflac, M, NOM mid-points of AD. BC mape Prove that CD and MN, when produce, are co/ Produce BR do meet C3 produced at.0. Join ON out fei DN cut AD at ht". In & ABN: • AM” #BN
Girts AM'
・응용
OCN:
AMD ANG
Gier
AD/BC.
ave find 용동 용문
AM MR.
NC (Given)
2. AM'=P" D
c. M' is the mit-pt. of AD..
By gubat.
Buf M is given as the mid-pt. of AD
At and Agr1 are concidunt pt...
O, M, N am
collinear
Example 3: ABC is an isos. 4 such that ̈28< < C = x «A,
Prove that A&*« B¢*+ AB- BC
Preet Produce BC to D
-B = LACE
➡α.+ <D
* 2 4D
RABAC
BACAD
* CD=AC
base da, isos.
ext. c. of a
Given
Hence, BR is a tangent of the circumcircle
of & ACD HA
In circle ACD
BD is
a
BA is a tangent and.
secant
BA = BC BD
= BC (BC+ CD),
BA* = BC + BC AB
tangent property
CD=AB.
Example 4: In 4ABC; AD, BE, CF are attitudes and #it is,
the orthocentie. Prove that
Prost
AH-H>= &H HE – CH-HF
H is the orhocentre of a ABC
« AEB ➡ ADB = 1 sha
2. A, E, D, B are concycliè,
In Circle AEDB.
AD. BE are too chords (** AH HD = 8H - ME
intersecting chords) Similarly, BCEF is oko a From which, we find Consequently.
cyclic quad
BH-ME='CH NE
AH-HD = BH HE – CH HE
A
con cyclic.
NOTE: The converse of the above example is also true.
then A. 8, D, E are If AH-HD = BH HE
AEB ZADE
(a)
Similary, BH›HE "CH_HE
Hence
« # C D 1 CAB
ADB - CFA
<, in some sey.
a cyclic guad
APDC is a sychic qund - (Al-H2=CH-HF)
< CFA = LADC
(a) and (by: << A?B=¿ADC and BDC - ÍS
AD LBC
Similarly, BEL AC and CFL AB
is the artkocerte of a ABC.
Example 5. ABCD is a cyclic quad. Prove that \· AC· BD = AB• CD4 BC-DA
Piect: Draw`«ADE = « CIB and let
DE meets AC ad E
<a>
«ADE »
DARD.
GADE VI a BDC
*** AD: AF « BD
A. AD-B¢= BD-RE
Since *. #BDO AECD°(AA...)}
<br
(a)+(b)
AB:8D EC: CD
ABCD Ba· Ec
AD-B¢ + A6 · CD = 6D ·HE + ¿Ð°EGA'
* &DÇA# + FC) - BD - AC.
NOTE: The converse in alte fut, and this is called the Holamy¿ Themem.'
HONGKONS
ar cacument, ushen produced, ao. also an alternative proof as shown in below: produced medis HM płoducat at 0.
CD produced mezis NM produced at oʻ, 4DÊN: AM IBN
ĐÓN
NƯỚNG
AM® MD and BM®NC (Gimn>
-
AM - MA
0 ́ ́und O' are both the pt. divides externally MN the same ratio. By the uniqueness of the pl.
O' must be concided with 0." Hinez BR, C'a'anată MN am concurrent, when produced, et o.
P is any point outside a circle., PR, PB niê two tangents from to the circle. PCD
of line cutting the circle at c and D. that the product of opposite sides of the
are equal.
m&c.AD
PA is a tangent
of circle ADAC,
(PACm 4 ADC" (4, 1)
SSAN A, PAC, PDA !
PACH ZADP... proved
APC:
LAPD
Common
APDA
AAA
岳路
Similarly,
est in a PDB
's
zore
quad
Crest, (a) mon
1 ABC is a triangle, a sh line cuts BC produced
RA, AB at P. Q‚R Hsp.; Cx in drawn par ée pa, meeting na at 24
BP: 80- BA
DBC: BP = BY – BR
BL
BP-BC: B?• ¿Î» 8× 1 &# CP : &P= XR4 BR.
4-#
Prove that
< Dividendo ›
# $ 4-4 ER AR-
ca 48.
48-42 - 88-**
Henée) B. C. AR-(-A)(--)
subst..
Note: This problem can also be proved either by drawing
`CK BAB to cut R&P at K, or by the perp.
AL, BM, CN from A, B, C
to PAR resp. The converse is also hue; and example 6 À
called the Menelans' Theryem,
31 unite ar
HINTS & ANS, To Ex, 16
1. Braws a met. of arez 31 2p, units (ruch as Lunit x 2x 15.5 m), by prob, 5 (this Ex.) & hav a square of equal area, and meosum its sides. 2, Draw A Bm G Cu-
Describe a segment- en AB A
celain an angle of se[cenbé 0)
Bisect AB at X.
ceube X, with radiu 55 cm
draws an arc to cut to atC.
Then ABC in the repaired .
- DEBAKKEIBUKC IN-DE 「鲩屬香港第三上午小學(大道東)
煒公佈。 本庭中文中基會考斯考科目,仍與往一科替代。 間舉行,其中無時間表,刻尙待最後决定一。惟今年仍有己公民一科考試,到下年 葳特定于六月七日至十八日之間善行營選|數學,化學,物現,寒酸,生物,聖靈〔 科目。至專門科目考試,則將在五月茶餐致),聖經(天主教」,佛學,術中學,獲導中學,亭流恩紀念中學,發業 彌事工,現正由有關方面如棄安神中。书·世界縢史,公民,地理,甲組敬學,乙組學,九臨三育中學,九耀艾光中學,大 〔轉鹅)本年度中文中學會灣各項囉 | 年相若,包括有中文,英文,中國歷史,一港九州公學,云江中學,賢繁中英文中 | 廿日至廿四日之間及六月十九至廿一日之一,音樂,工業繪圖,縫衣, 傳記,打字等中學,崇淑英文中學中文部,碎堂中學,人如長有上述資格,並可於本年五月六日 坐起則將取消而欲爲「柽濟及公嗽實務」寶書院,南寶中華,新中文中學九分份,信封面寫明「願役」字嵗,投驛本港 校,貝爾暫院,東方中學,母書院,‘管仁街太華三院感務部爲荷」。 信中學,瑞利諾神父教會學校,利校投無者,於五月二日前栖只愛一
四十三月四年八六九一腊公年七十五
中藥 特
考普通科 五月廿日至廿四日 六月十
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+
中文中學畢業會考
與考學校八十八間
***
+11E-HEEK-
育
〔甲】香港區鯤 校 低院學校,中華基督甄年會中學,中正 ,茲解恣院,襪英中壓,至誠中學,香港。(股) ),漢莎中華,香港三育中學,體劑中學,西鍵樂實中學(密雲)長洲官立中壘等 ,店橞中藥-殯璠中學•官立夜中學(蕙濱 中學(屏山),三臂害院(清水灣 ) 學,成人中文夜中泰〔地區),仿林中,朗個藝中學,鳥溪沙兒童新村中學,柏丽 中學,商中學,令空蒸中學,孔裂豐中賚中學(大埔),粉靬心盆中學(粉嶺) ,鳳溪中學(上水,元朗公立中學,元
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| - LERULEØRNEKELLE KERTE - ES54 - TRE- 一 生外,其餘均係來發八十八間中文中學撒聖瑪利書院,逸仙中學,將做中學, 一有八千五百名,其中有一千四百名自學,培知中華,培聖中學,慕中藥,高 4年參加中文中學售者之學生,一共巴富爾官立中學,培琪女子中學,培正中 學校,不舍液學校,崇正中學,東華三
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|-*2##-##+2' NTE34
新法中學,獲女子融業中學,青血女子
東華三院小學
急聘英文教師
8
(i)
DraisTM
佛敦大
GOR LA& (euf
, Jain @p, cut 00 at T
> Draw C°C # LAB,
CO alQR
úra Join OT and produce it to meet PCC' of C. (V) Centre C. radius (P(aet) draw a circle (.one sol.) Replace or by R, T' then... we find the other solution.
6.4, draw sg. ABED of ama 12 yp units. semicircle EÃƑ, «¿¿EF* ? units.
Ab'm one side of of. ABCD. 3°F = x. Efound by "meessurements {
Complete rect. BG
*** Rect. £G=
But FF=4,
A'C'm it) *7.
•. £3·3°F ={4's') = Sq. A
In giden rect., if AD >AB
Ak".
is Complate ADCB to be a sf. ADKH.
Vy with alta. Alit, describe a semicircle
on AH
E
Hi, Prodice CB to meet the semiciak att.
iv, Jom RE
v, Describe a squat în AE,
Then, AEFG is the requind square.
J
*
NOTE This Construction, is based on lythagoras,
Since
But
H HEA = 1 ula
- in semicircle. A HÀẻ = A DAE SAS « HAQ > > ERG =± 13. AF 4 DAE = «DAB = §. rect AC,
*g. AF = rest. AC [125 An alterative construchio, 1
¤, Prodaœ AB to X x4 BX=BC. 1, with dhà. AX, draws a
semisvele.
6, Produce CB » meet the
semicirch at E
4, Describe a of BF on BE.
Then B&F6 "is the requised square.
D
This construction depends on either the property of intersecano chords or the mean proportional.
CARRCISE. 22
1, 370 is the incente of aABC,
ac at D, then AD : OD = A8+ AC + BC.
2, ABCDEF is a
Produce 10 to cut
hexagon, inscribed in a
circle.
If the diagonals AD, BE, CF are concrement at P, thes to $.4
3, Two cireres met at P, Q. C is
Draw st. line through a cuts the circles of A, D 4 pont m PQ . and B. È resp. Prove that 18:10=ED IDC. 4. If, in fig an shawn, P is any point inside a ABC. AIX,
BPY, LPR an st lines
rove that the CX-1