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TRIANGLES H Basic Proportionality theorum C8PT> Tl a lin is drown parallel te one Side of a “riage wrersecting he othef tivo Sides in distinct pants then “the sther two sides are divided in the same ratio: 4) 'y) £ (0 Ao = AE Q) FO = Ae 06 ec Ab AC b t @) eo = be cA #LP Dell Ac. Find ~ mam = “At3 f A*\ aS a /\ax% 0 € =) we = WE eM +3 a+\ ars HUE PQ ibc. e = 4 an AC= 204. RW AQ (> \e) ; \ Un+ An =20.% p \ Mn = 20-4 g cq HR = (2 Ag = AP = 4.6 Gn ¥ Converse of 6PT 7 TL a line divides any two side of 2 traps iW the Some patia , then the line iS parallel te +ne third side. #0 Ao = bem , 0b = Sam , AE = Ban , EC = (2cm CLADE = WB" Fl Gre » § = 2 4 3 2 7) (a . 0€ Il Be Coonverse f PT “+ CAec = 4B C Corresponding Argsier ) In AROC An = AL — NO LC © From @ avd W, we get AM = Ax — Mé ve * Pron} o Get? Given 7 AAG H€ Ii BC To Prove 7 pp = AE 0& ec Cowstruction 7 Soin GE Draw EF oC Prov, > r.(AADE) = 1 x AD xee 2 Dr. CABDE) =1 xepx LF F) =F f * AD xe 2 wo — (I) x « Boxee Ar. Laave) = 1x AE k OG Q Ar. CAME) < 4 x EC KX OC 9 Xx AC on =) pe —((l) Z x EC x pa EC TL we equate LHS then AHS will automation, be equot » Ar CAROL” = Ar (Anoe)” Ar. CADE) Ar CACOE ) WA S fr. (ABDE) = Dr CACDE) — Cr. 0 As ow [I lines is equal So now KHS is also equod - “. AO = AE 06 EC HLP AB (1 CQ and AC Wl PR. Show that BCIIQR ? In apog, oR = o& —() Cé6pT) aN Re 6Q AOS Q .< In APoR, on = 0C —() C8ReTD re CA From © and () 06 = oC /{ BPT in AROQ BQ ce by Converse 4, BPT, Bc (1 Qk Fle PSCD is Q trapezium iw which Ab I] Dc, diagvals intersect Ot pont O. Shoo AO = 20 DO In Apoc 5 RE = Ao — (1) Eo oc Dp © In A bro, ne = og — © €0 00 From 0 and W, fo = 08 ot. 00 S AOXOD= 08 XO > Ado = co Bo CO FUP I, 3 Il tines dL, m,n an intersected lby rrans versal q and s OS Shown In the biguee. Shoo AG = DE In LACE, S 7\9 € 6 = 90 —@ ia &C oF “LOAF No» O€ || OC and OF (( AB -"- AR Cd ® Siorilariti > Shoge 8 Some , Size mow differ: OB = 4 22 Be : =2 @ ac = ee PR = Ab = 6c = AL ?Q QR PR 2. L\RGC~ APQK Hue Ask ~ Aoge , LROQ = RAS ond LORS = To. Find Low Ond LOQP- Zsoe = Qs’ (Cvoay 250 +2 = 360 * = 5S (N “* LOSR = Loge = SS” HL OA x 06 = OC KOD . Show thet ZH = LC ad LQ=LD OR = 6 — () F 6 LAod = Leos Cvond —() 2 From O and ©, LZ\ pod ~ Alob by SAS Ry CPST, ZA = LC Ze= LO = (peg = Zpec | cAQP = Lace ". A APE ~ Apec ee = (R = AQ Dt = CQ = LF AG ac Ac 3 3.6 YT 2 > |! = £Q > ( = 36 2 PQ = 1,2 Cry 3 3-6 2 HL MNO? is a porallelegram and Ab | MP. Prove Qc \( Ro ZQu? = LQns Crorvesponding a) Laem = Lash C Corresponding 1 by BAA Simi tariti, . Lars A Qe ~ Agaé In Ache ZCNO =ZCAR C Cosstspording *) ZAC = LACB C Commen) » By ANA Similarrty, D CAE ~ AlCNo From AQMP ~ Aare , Qm = Me = Qe ar AS ae —0 Frm ACND ~ ACHE CN = NO = CO ——. CA Ag Ce 6 i From ( ond (WW OP = ( "" NO= MP as oppaste “ae Q6 cb of pasalleagram ore equal f from thig, we Can Saw nok QC (| PO C by Converse 6 BPT _) = = h = => HL LACB = LOA, AC= bum, AO= Bem, BO=? C y LACe = Leon Cavern) 8 L6= 46 C Common ) y, 4-340 6 -- By AA Similarity, AADC =~ ARC -+ by CPST, po =Oc = fC J 3 = g AC C® AS 8 Ae L T SAB = C4 D> fO= SH 3 > AO+ BO = AB 2+ B80 = 64 > &p = ey -3" » 64 -3 3 3 3 6D = 55 
