日僑華
二期星三日二廿月四年九六九一层公年八十五國民華中
TY BO THE TON
1969 歴文中學専
題預習
真二第張六第 日六初月三年商已變
WAH KIU YAT
1969
C英文中學會考試題預習
數學科 (廿五)
ACBD= BC
飲陽絡文
Đemple Lo AEG is an Laos, a such that h
Prove that ÁBB=BC2+AB-BC
MATHEMATICS (25):
LESSON 251 SIMILARITY
IMPORTANT THEOREMSI
1 Theorems on the ratios of a" a formed by a #line:
(1) If a st. line is drawn parallel to one side of a -
then it divides the other two sides, produced if necessary, proportionally.
123 If a et. line is drawn to divide two sides of #M
in the same ratio, both internally or both extern- ally, then the st. line is parallel to the third- aide.
If AB and CD are transversals of a set of parallel lines, then the intercepts made by the parallel lines on AB and on CD are in the same ratio.
(4) As shown in fig. in which XT # C
then, (1) 4 -✨
般
A
X
B) Theorems on the ratios formed by angla-bisectors:
(1) The internal angle-bisector of a ▲ divides
(internally) the opposite side in the ratio oi the other sides of the triangle, (And also the converse.)
(2) The external afgle-bisector of a 4 divides
(externally) the opposite side in the ratio or the other sides of the triangle.
(And, the converse is also true.
C) Theorems on similarity of a
(1) Equiangular As (are similar)
2) three sides (of As are correspondingly) proport-
ional (then, thes are similar) (3) ratio of 2 sides, inc. < (then, the
eimilar):
As are
(4) If two e ars of equal attitudes (or bases), ther
the ratio of two corresponding sidess
15) The ratio of perimeters of two similar
4:3.15
equal to the ratio of two corresponding sides. (6) The ratio of areas of two similars is equa)
to the square of the ratio of two correspondin sides.
Theorems on similarity of circle-
(1) intersecting chords,
(2) tangent property.
If at Lines AB, CD are divided (both internally or both externally) at X such that XA XB = XC-XD. then the points A, B, C, D are concyalic. (4) If AB is divided externally at X, and if C 15
pt. not on AB, such that. XA XB XC Then, the circle ABC touches XO at C,
AMPLES
Trapezium ABCD with AD // BC, MN are the mid- points of AD, BC resp. Prove that. BA, and I, when produce, are concurrent Proof Froduce BA to meet CD produce at 0
Join ON and let ON aut AD. at 'M' In AABN: AK // BN
AM BN
= 응용
AOCN: K'O // NG
rrooi: Produce BC to D s.t. CD-AC-
BACB base saj isos a.
ma,+&D ext. z of
2 D
ZB2BAC even
Hence, BA is a tangent of the circumcircle
of ACD at A.
In circle ACD, BA is a tangent and
Example
BD is a secant
BABC-BD-
BC (BCCD);
BA*— EC*+ BC÷AB
tangent property
CD AB
In A AD, BB, CF are altitudes and H 1s the orthocentre, Prove that.
AH-HDBH HE➡CH•AP
Proof: H is the orthocentre or AABC
LAEB - ZADB =1 st.4
4, E, D, B are concyclic
In circle AEDE,
AD, BE are two chords
AH HD BH HE
(intersecting choras)
Similarly, BCEF is also a cyclic quad. From which, we find BH HECH AF Canggquénti AH HD-BH-HE-CH-HF-
NOTE: The converse of the above example is also true.
If AH-HDBH-HE then A, B, D, E are concyclic.
2AE8➡ADE
s in game seg.
BES CFB
Similarly, BH HECH HF BCEF is a cyclic
s in same seg. supp. a
(a)orzBEA = c CFA
Hence ADB CPA (AB) But AFDG is a cyclic quad.
(b) CFAADG
(AH HD CH HF)
sin same seg..
From (a) and (b): ZADB=<ADC and BDC is a st.line
\*. ADLBC
Similarly, BELAC and GFAB
H is the orthocentre of ABC.
Example 54 ABCD is a cyclic quad. Prove that
AC BD-AB-CD+BC-DA
TOO:Draw ADODB and let
DE meets AG at Els
ZADE=< BDC- Cast
DAEDBC
MADE
ADAD
(a) LG. AD BC
K'S IN SORE:
BDC
A.A.
BD: BC.
BD AE ·
Since ABD - CD (A.A.)
AB:BDEC: CD
(b) or AB-CD BD-EC
(a)+(b): AD BC +AB• CD=BD- AE +BD-EC
ED (AE+BOY #BD-AC
NOTE: The converse is also true and this is called.
the Ptolemy's Theorem,
Excamole 6: ABC is a triangle; a st. line cuts BC produced, CA, AB at P, Q, R reap.; CX is drawn parallel to PQ, meeting AB at X. Próve that
(a)
BR
BP
QA
英文科 (廿五)
INGLISH (25)
Answers to Faber KXIV
Sne arriveu hume: so ima ner nevner, a large,
hot, untidy woman, doing some of her hurried. cooking. The fact that it was a warm day and wOULE be a close evening had not deterred Mrs. Salter from attempting a very hot and heavy meal, for she took no notice of the weather but catered by immediate inspiration, so that in January she might offer you cold pork pie and tinned pineapple, în. August fried steak and a suet pudding. This seemed reasonable to all the family except kust, who was, they said, a bit hard to please.
"Mother," she announced at once.
"Yes, love?"··
"I've got the sack..
"Well, you ailly girly cried Hre. Dalter, lifting her large round red face from the low oven. "What ever for?!!
Rose told the story or une arternoon as well as she could, blaming neither herself nor har employers. Her mother did not clearly understand what it was all about, but by this time she had
· given up trying to understand all these things. Every other day one of her children brought home some fantastic tale. But she was neither cross nor worried. She was even pleased. This was drama of a kind; and the Salters liked drama.
(1) on
(11) of
(2) fróm
(12) fo
(3) on
(13) to
(4) of
(24) in
(5) for
(15) in
(6) on
(16) of
for
(17) at
(8) of (9) in
(18) or
(19) in
•(20) of
10) at
C. (1) I do not think he will do anything
remarkabla:
(ii) I have received your letter.
(iii) He nearly succeeded."
(iv) The only way to treat such a person 18 to
leave him along to ostracize him.
(v) The defenders still held out, clinging to a
mere possibility that the king would send help.
(vi) It was evident from their noisy behaviour
that they had drunk too much wine. (vii) When the roll was called, Johnson wab
noticed to be absent.
(viii) After a long day in the open air, we slept
Boundly..
(ax) Unless you are more careful, you will deval
be scolded (or reprimanded).
The boy received reasonable wages for nis
services.
- 응
OBC, AD // BC, we find
AME
EN=NC (Given)
** AM'➡ MED
1.6. W' is the mid-pt. of AD
But is given as the mid-pt. of A
Mpand Ware coincident pt.
DN are collinear
1, NB, CD are cóncurrent, when produced, at 0
hava also an alternative proof as
below's
Let BA produced meets NM produced at GD produced meets MM produced at
IGA OBN AMBN
In AO'CN; HD // NG......
AMMD and BN- NC
ON
(Given)
1,0. 0 and 0 ́ are both the pt. divides externally UN at the same ratio. By the uniqueness of the pt. 0 must be coincided with 0. Hance BA, CD and MN are concurrant. when produced. at 0.
Example 2: P is any point outside a circle, PA, PB are
two tangents from P to the circle, PCD la s at. line cutting the circle at C and D. Prove that the product of opposite sides the quad, ABCD are equal,
To Prove: AC•BDBC AD
Proof: PA is a tangent
of circle ADBOA
PAC ADC. in att seg
Consider As FAC, PDA:
PAC ADP proved *APC = APD
common
APAC PDA AAA
AC
PA PD
Similarly, 4 PBC 25 ▲ PDB
FA, FB are tangents from ♬ to circle. ADEC.
length of tangent.
PAPB
口味
Pro
(a) From 4BPR;" CX # PR
BP:BC=BR:BX-
BG:BPBX BR
BP-BC:BP-BR-BX:ER"
(Dividendo)
CP BP
XR: BR
ВР
BR
XR
QA
b) BP ca. AR = BR SA AB
AR RB
Subst
In A ACX
QA XR
RQ // XC
AR
AQ
Hence, BP. ca AR = (RX) (AR) subst:
NOTE: This problem can also be proved either by drawing
CK// AB to cut RQP at K, or by the perp. AL, BL, CN from A, B, C to POR resp. The converse is also true; and example 6 is called the Menelan Theorem."
HINTS & ANS. TO EX
1) Take point E on a such that
Draw EFAQ. Then, REFF is a square. Flies on the bisector of REF If Q, R may lie on the BA, CA produced, then, we find 4 such internal / bisectors.
The prep. bisector or AB.
3) (a) A circle, centre A, radius PA,
(b) Surface of a sphere, centre A, radius PA. Draw a cone to circumscribe the two spheres. The vertex of this cone is a fixed point. By rolling this cone on a plane, we find a group of concentric,
未完轉入第六張第三頁)
circles. Hence, the required loci ane two circles.
5) The two arcs as shown (or two
equal circles) together form
the locus of vertex A
*- Z BIC =180 −1( ZB+/0) −90 → → And Afixed. /BIC const." Pt. I will approach to B when A tenda to B.
Arc BIC is part of Locus. (Similar reasonings for are BIC) Ans: The locus of I is arcs BIG & BI'C
6) Two methods:
By calculation:
bisector locus).
b) by bisector locus.
A A
BAN =/5"
EXERCISE 25
1) If 0 is the incentre of 4ABC. Produce AO to cut
BC at D. than AO:ODAB+ AC
BG
2) ABCDEP is a hexagon, inscribed in a circle. If the diegonala AD, BE, CF are concurrent at P. then
BC DE FA
3) Two circles meet at P, Q. € lo a point on PQ,
Draw st line through C cuts the
and B, E resp. Prove that
AB: cles at A,
Ir, in fig. as shown, P is any point inside, ABC, APX, BPY, CPZ are st. lines. Prove that
AZ BX CY
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