1. sum of 
Adj. s at st. line
Adj. s supp.
s at a pt.
Vert.  opp.  s
corr. s eq
corr. s // lines
alt. s eq
alt. s // lines
int. s supp
int. s // lines
ext. s //
sides opp. eq s
base s // isos 
Transitive property of parallel lines
Corr. sides,   s
S.S.S., S.A.S, A.S.A, A.A.S, R.H.S
Corr. s,   s
A.A, A.A,A 3 sides proportional ratio of 2 sides, inc. 
Corr. sides,  s
corr.  s,  s

2. If x + y = 1800 then AB is a straight line (adj.  s supp.)

3. ISOSCELES TRIANGLE A A x x y y B C B C
If AB=AC then x=y (base s, isos. )
If x=y then AB=AC (sides opp.eq s)

4. Example 4
Line segment AE and BD intersect at C. AB=ED and BAC = DEC . Prove that AE and BD bisect each other. A B C D E
In ABC and EDC,
AB=ED (given)
BAC =  DEC (given)
ACB = ECD (vert.  opp.s)
ABC EDC (A.A.S)
AC=EC (corr.sides, s)
BC=DC (corr.sides, s)
AE and BD bisect each other

5. Classwork 5.4
1. AB=AC and ADBC . Prove that BD=CD.
In ABD and ACD, A C B D
AB=AC (given)
AD=AD (common side)
ADB =ADC=90 (given)
ABC EDC (R.H.S)
BD=CD (corr.sides, s)

6. Classwork 5.4
1. PT and QS are straight lines. They intersect at R, PR=T and RPQ=RTS. Prove that R is the mid-point of QS
PR=TR (given) Q P R S T
RPQ =RTS (given)
PRT and QRS are straight lines
PRQ =TRS (vert. opp.  s)
PRQ TRS (A.S.A)
RQ=RS (corr.sides, s)
R is mid pt of QS

7. Example 6
BD=cd, DBAB and DCAC. Prove that DC bisects BDC B D C A
In ABD and ACD,
BD=CD (given)
AD=AD (common side)
DBA= DCA=900 (given)
ABD ACD (R.H.S)
BDA= CDA (corr.sides, s)
DA bisect BDC

8. Example 7
AC=AD, BC=BD, AEB and CED are straight lines. Prove that ABCD.
In ABD and ABD, C B A D E
AC=AD (given)
BC=BD (given)
AB=AB (common side)
ABC ABD (S.S.S)
CAB= DAB (corr.sides, s)
In ABD and ABD,
AC=AD (given)
AE=AE (common side)
CAE= DAE (proved)
ACE ADE (S.A.S)
AEC= AED (corr.sides, s)
But AEC+AED=180 (adj.  s on st. line)
AEC+AED=90
ABCD

9. Example 8
AB=AC, b=c, Prove that BD=CD. A C D B b c
Join BC, such that b=b1+b2 and c=c1+c2
b1=c1 (base s.isos )
b=c (given)
b-b1=c-c1
b2=c2 A C D B b1 c2 c1
BD=CD (side opp. eq. s)