1966-04-05 — Page 22

買二弟張六第

日五十月三年午丙春夏

WAH KIU YAT PO

華僑教育

***A

試題預習專欄

數學科

( 十四 )

歐陽鋊文

MATHEMAILSSS(4)

LESSON YA CIRCLES

Example 1 Con chords of a circle)

X. Y are the mid-pt of the chords AB CP of a circle, centre 0; XN. YM are the & from X. Y to CD. AB resp. If xx cuts

YM at P, prove that of

XY bisect each other

Proof: Join ox, OY

*.*

M

A

x is the mid-point of the chord AB

. OX LAB

But, given YM LAB

Similarly, both of XN are

2. OX HYM

CD

OY HIN

'. OX HPY * OY #IP

PYOX is therefore a igram OP XY bisect each other.

By definition,

Example 3 ( Angle Properties of a circle ›

The bisectas of the angles ABC, ACB of a ABC intersect

at I and Cut AC, Að

at Y. Z resp. the

circles 812, CIY

期星

日五月四年六六九一层公年五十五國民中

育教童

Example 3:

ABC 13

an

equilateral s

inscribed in a circle

P is any point

arc BC

Prove that

PA = PB + PC

Loof Produce BP to D such that PD=PC

In ACPD

PC-PD

• CPD = BAC = 60*

Conso

K

Ext. of cy clie quad

Consider < APC. ACE

AC - BC <ACP = B C D (~60 + x)

СР- Со

SAP ESC D

... Apa BD

Hence, with–cente 0. Aadics Ok=Ź(P-8); we

can draw another circle to touch each of the

given circles.

HINTS & AUS. TO EXTA

and

The tima. laken

Oct. 15-y

! Since the cost of bill is equivalent to the Princ, in S. 5.

the banker's discount is equi. to the Interest

Sept. cao (of the same vi.) * grace's days. - 28 days 3 yr.

2

23 The bankers, dù count. =6% d£ £389) × 365

Ok...since I=PRT =

x% xx

Sides of a PCD.

5.AS

NL GAYS T = 3656ays

3£{px2n1] x[ 14 £*£% + 1)÷ 10* which is what we called the "THIRD - TENTH -- TANTH - ROLE" Discount <= £ (3 8 5 × 2 × < 774)× (1+ 4+*+20=105

Since

BDBP + PD. – BP + FC

PA = PB + PL

Note: you may take a point E on

AP 5.1 PE- PB: Then, prove Mat A8PC = ABE (mote · A BPE À equi), then

PC = AE..

~LLIS

; by practice mrkod * 3 s id.

217

Example 6 - Tangents)

The truckTM present worth The truce discount

The time taken

As shown in fig. A,E

are the centres of two circles touching the parallel lines GOH.

I

The rate = PT

3,

EPF; P Q touches

each circle Prove

that (*) APBQ is a recr.

b) PQ = AB.

freef

Y

AP bisects

Since PE, PQ are tangents from P to 0.A

Similarly

Bit

e a black LEPA = PQ H

EPQ

Pan

GH JEF.

zee... P

=

p = £786 -£30 = 750.

=I= 30.

- T = &mth, m gr.

67%. p.a.

ket to be the principal then tax bɩ the amanRT as At Simple interest i Interest = 221-2x = £*X 57XT

.. 7 = 20

b. At Compound interest: Amount - £ax = £x ( 1 + 5%)*

.4, (a)

Subst. 2=

= {(ag-1) inh_*** 517 +35*+340. then домогу навалунах (урт) =,

ANS. (-3,-4) 8+ (-5-7) (b) From (2+ y) −x*+ y * ~ 6 (X-3)*=0

we have [(x+9)}+=(x-9)][(x+3)=3(x-5)] = Hence either 31-4=0 o zy-x=0,

Ans. (-1£ ~*) or (2, 1)

meet again at x B Prove that

zYXZ+ 481£= aƒ«.

Proof Join IX. with the notations as shown in fig

<IXY = C,

Gar

some segner

112 - 6,

Since

<1 = C

and by - B2

given

120

In

A IBC

ZX Y + - IA & =\Cy + bz

< BJC + but

Similarly. AQ, BP are the bisectos et

the alt. <: GQP,QPF

...AQ #BP

40.1 BP

AP Ba is a Hgram

AQ. BQ are bisectons of two adj. as

Example 7 (Alr. segment,

(cirere

and be when

ABCD is a minor are of

such that AB= BC · AS

produced meet of P. and DA produced `to meet the tangent B1 at T. Prove that TP=TA Prack:

* AB - BC

give m

stand on equal chods are equat.

Example 2

( Concyclic Points >

ABCD is a ligrans

o à a point in side

ABCD such that

-A0B +« COD=Ara

Prove that

4 060 - <ODC.

Proof: Draw APADO,

ONG KO

sides of amp are parallel to

Since the D

the three

"But

Sides

of a coo

AB = CD

AMPE ADCO

Dep siles, ligram.

AS.A

APB Doc

are Concyclic

Join OP.

OC LPB

DP # BC

Constr. Place of

But

Example + ( Chords & Arc

DABC inscribed in a

circle as shown.

mid - point of mina

are BC. D is a point on DA

DO=DC. Prove that ou

incentre of ABC

Join OC

AT

on a st. line.

shay of rest.

Let

sii all.

+

• TPA = 22-

same

segm

-TPA = x (« da)

TATP

sides øp, equal –

in same segn

Example & ( Conface of circles)

InaĦBC, AB=Þ, AC=8

<BAC -90. and p >8.

O is the mid-point of BC Circles are drawn with 45 and AC as diameters Prove that two circles can be drawn with o

the same circle, igual ater subtend epi

sector of 2 BAC

DAC:

from LA

conclude that

ISOKA

the dist. P Q = 1 mi.

spead = 2 mph & Yé speed they meet in

mph == mph

the after X starts. That is.

[the after & Hence,

I¤ ̧ 21(47 −3)

20

ANS, A+ Neo

Xf distance + Ys def. == which is indep. with

Y yo a the inner radius of the path.

I yd. the width of the path.

area of path = m[I+x)" - my*«@#!x+7x* cost for graveling the path = &(2nrx+1X3] 8h.

= 6

12.9.

Length of has edges =27 [k+7+x]=ancer+x] Cost for edging by stone = 21 (artx)· 3 sh=219_165.

Exercise 7 A

ANS. I loyd X= { »

A triangle ABC IS

sectors of the Is Stibed in a circle, and the bi-

the Off at X, Y, Z... show that the as of oxyz are resp. 90-£, Ja-m

Two Circles meer at A and B as shown AC, AD are the diameters of each

Plove that

are collinear.

"wo circles meet at 4,8

Prove that

CD is a common fangent

AABC IS Mscribed in

as centre to touch each of 4 their radiizin terms of t

circles - and find

of a ABC. The three

cides:

PROOF

Let E, F be the mid-pl. of PB, AC resp.

Join of and pro

JOIN OF

where OE-

••-OH =±(ABTAC)

Similarty 10M = 0F +FM:

Where OF = $482&»FM=FA®£AC,

OM START AC)

From (a) & (b), we have

DH = OM

Hence, with cemte 0, racine, OH= £(3+8), AL

CONSIDER WIE

OK FEK - FO

==(AB-AC)

CAB-AC)

the

produced to meet the circle again Kresp. Prove that: O is also the

Two equal chords HB, CD meet, when

Prove that BS = D6.

at 6

ABCD is a cyclic quad,

of four circles ins

Prove that MNPQ IS

Centres

ABD. ALD. BED.

The diagonals of cyclic quas. ABCD cut each other

af yt = at P. Puve that the £ from P to BC

when

pindisced AD C Brahmegupta Theorems

AOB COD AIR

I diameters of

Two chords CA. ca cut AB at that H K. a. P are concy clic