1969-03-25 — Page 23

REFES

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HR-BRIZURAY MOND KOO• ***E-RICHARD HUM BER-IKEBUSKLHERZB-GES 唱片說明突無書及所學會共合照。

地出售得更多瞭學基金,以英技术沿黹年。

日但踢起向入五交之路。 HEROZEBKEE

語

二期星百五廿月三年九六九一层岳年八十五层民華中

截止日期各不同

聯合最早:崇基最後

師特在續

黃海

CEGF

PIM GO

line and

M

collin car

And

EFG #AMOG

(AB+

provet

OG = GF (-SH) sides of so

·GH = 2. GO

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1969 FU?

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數學科

歐陽鐇女

D'FAC).

DAC=t<BAC

|_ MATHEMATICS (21)

LESSON 21% Inequalities & Triangles

(Including the 5 dentres)

1. Important Theorems for reference

Lo prova: DE bisects BC

A) For congruence of triangles:

(1) SAS 【2】ASA

(3) AAS

(4)999 (5)RHS

Incamla 31. In ▲ ABC, ABY AC, take points 0,5 on

AB, 4G produced reep, such that AD=AL

AC). Prove that DE bisects BC.

Given: A ABG with AB > AC

AD = AE = λ(AB+ AC)

Proof: Let DE cuts Bc. at P

Draw CF// AB: anak

B) For isosceles triangles:

1 r base me, iso8.5% (are squealy

23 sides opp. equal «; (are equal)

: bisector of vertice bisects also the base at

rt. ms.(The converse is also true) bisector of the ext. 2 of the vert. parallel to the base.

6) For right-angled triangle:

*It is always equidistant from the mid-point

the hypotenuse to the thres vertices,

6 Hypotenuse is the greatest side in art, a 7: In a rt. with one acute – equals 60° (or 30°)

the hypotenuse equals twice the shortest side:

For any triangle (on inequalitiss);

8:41 Ext of a ▲ > any, int, oppuss

Any two sides together > the third side,

10 The difference between any two sides the

third side.

11 i The greater side has the greater opp.

(The converse is also true)

E) The rive centres:

12) The L bisectors of the three 82068 01 a

concurrent. (Circumcentre throrem). The circum- centre is equidistant from the three vertices. 13) The bisectors of the int. 2, of a ▲ are con-

current. (In-centre theorem). The in-centre is squidistant from the three sides.

The bisector of one int. and two other ext. bisectors are concurrent. (Ex- centre theorem) The ex-centre is equidistant from the three sides.

The three medians of a are concurrent.(Cent roid theorem). The controid is + of the way along each median measured towards the vertex, (5) The three altitudes of a ▲ are concurrent.

(Orthocentre theorem),

(a) The à whose vertices are the feet of the altitudes is called the PEDAL TRIANGLE OF the original

let

CF CHE DE at F.

AD = RE ➡CAB+AC)

given

ABT AC = AD+AE AB-AD➡ AE - AC

BD CE

ADm Af" and "BD= CF

CF CE

DBFC

is a #gmm. 6P= PC

ming. Ilgram

DE bisects BC

Example 41. In ABC, ABAC, BC, CE are the bisectore

·B,

rasp.; prove that BD

Proof : The proof is given here in

outline form.

(1) 2-ACE → ABD

Draw CX

such that BCK = ABD Then BOX > GBX

BX > CX

Take 1 on BX

CE.

BM-GX (V Lies between

B and I). Draw MN// CX (N lies between B and D) Then ACEX

BUN (ASA)

* GB == BMC BN ED and BN➡ CS

EDCE.

Prample 5 Prove that the distance of the orthosentri

of a triangle from a vertex is twice the dis

thee of the circumcentre from the opposite

side.

ABC

Proof; Let OH be the circumcentr

and orthocentre resp, and let M, N be the mid-pts. of HA,

In ABC

Join UN, PQ.

BQ=QM

and

If the A is acute-angled, the ortnocentre lies inside the ; the orthocentre lies outside the ▲ for an obtuse-angled 4 lies on the vertex of the rt. 4 for art. A. c) the centre of any import

ant circle associate with the

The

HOLESTE

word "centre" is often used to denote the common point of intersection of three or

more: concurrent lines,

If the 4 in equllateral, then, the circumcentre Incentre, orthocentre and centroid all coincide with each other.

Conversely, if any two of the circumcentre, in- centre, orthosentre and centroid coincide one. another, then the triangle is equilateral.

2 EXAMPLES ↑

Example 1. In rt. ABC, 1 is the mid-point of the

hypotenuse BC. Draw DLBC such that D lies opposite sides of A and that DBC Prove that AD bisecta 2 BAC,

Given: ABC with 2 Aw❘rt,«

1DLBC

To Prova: AD bisects BAG

Proof! Join AM and atrava AM-1[MD

MD 2BC *AN LBC

M is the mid-paint of BC.

OQ LAB

-resp.

"Given

BP = PC

- paint

IN HEAC

AP 1BC and ADLBC

the sides the inta's are

Simihily. He ==

UMN GM Ach y aquat

AJA

OP= HM (={HA)

Or the problem-mvæ íso be proved as follows:

With centre 0, radius 08, draw the circumcircle ABC. Produce BO to meet the circumcircle.

(1) AHCX is a #gram

CX AB

(11) ▲ BOX is rt. 2 at G and OP=»C2

le 61 If AD,BB, CF are medians of ABC, prova that AD+BE+C? > → (AB+BC+CA). Prooft Let G be the centroid of ABC From: 4 GBC 1 GB + GC > BC\\\

"Similarly, by

MAKME »M

base 45

att.

LAN

Z • GAC

TAMB

ANC

proved ment of ec west of

RMAS

·BAD="CAD": AD. bisects. ZBAC

Example 2: In rt. ABC4C

Prove that <DAGE†2 BAC Proof: Let N be the mid-pt. of 199)

Jain BN

In rt, a DBE, N is the mid-pt. of hypotenuse DE.

NB = ND = NE

BA=&N(m£#8)

BC / AC, ED = 2

CF +

SCA †AD+ZBE >AB

A

AD + BE+ CF) > AB+&C+ CA

A0+ 8E+ CF > 4( AB+8C+ CA)

Example 7: If H, U, 0. are reap, the orthocentre,

centroid and circumcentre of a ABC; that

(a) H, G, D are collinear, (b) HG 2.GO

Proof: Lot AD, CR be two altitudes

1, be mid-pt. of BG,AB resp

GA GH

by mid=pt, theorem

A GHA, by

EF HA

From Example 1; i

EF

Prove

OM BEF

MOG

SAS

NOTE OGH is called the "Euler Line" of A ABC.

Example 8 x,y,z are the mid-pt. of the sides of

ABC; prove that the orthodentre of "A 112 la the circumcentre of ▲ ABG Proof: Let 0 be the orthocentre of

Join XO and produce it to cut IZ at P

are mid-poi

of AB

AB

circum centre of AABC

Similarly oy Hence

AC at

HINTS & AM

Ex. 20

AC

CB CA-sp

N1AB

VMD

MAND

«NDC = «C»«HMDT 4MND

=DN=★AC

Produce CT to mast AB at R

T=AACT (ASA)

TC

BDM DC DT # CA

Am RAB -RDA - LOAD

ZBAC

act if «B dy

=* (BA- AC),

BMH (ASA)

me- points HKIMN KL HRIBC. EXERCISE 21

1) In 4 ABC, AB>AL, AM 11 8 Medlar. Prove that.

BAM

2) I AD, BE, C are the three medians of a ABC

then 2(AD+B5+ (F) > AB+ BC + CA > AD +BE+CF.

In fig. as shown, the bisector of

BAG meats at P. the line which bisects BC at rt.cs; FX, PI the perpendiculară from ♬ to

AB, AC Find in the ring

three pairs of congruent angles and prove the congri-

Prove also that · AL=4(AB+ A ).

A

I 0, 0, 5, rare resp. the 1-centre, and ex centres of ABC, then 0 is also the orthocentre

A DIT

are four points on the circumference of a circle. Frove that the perpendicular bisect- Ora of AB, AG, AD, BC, BD, CD are concurrent.

AHUJE Auru are squares on AB, AC outside ▲ ABC. Let AHBC Frove that AH, BF and CD are

(ent.

concurr