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|_ 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