sum of  Adj. s at st. line Adj. s supp. s at a pt.

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Presentation transcript:

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

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

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)

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

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)

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

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

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

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)