[Citat] 1. ABCD paralelogram, P apartine lui AB, PC intersecteaza BD in M, PD intersecteaza AC in N, notam cu O intersectia diagonalelor paralelogramului. Aratati ca OM/MD+ON/NC=1/2. |
ABCD paralelogram: AB/CD = 1
1) tr. APN asemenea tr. CDN (AP||CD) = > AP/CD = AN/NC; AP = AB - PB
2) tr. PBM asemenea tr. CDM (PB || CD) = > PB/CD = BM/MD; BM = BD - MD
3) Adun, membru cu membru, egalitatile 1) si 2) = > AP/CD + PB/CD = AN/NC + BM/MD = > (AB - PB)/CD + PB/CD = AN/NC + BM / MD = >
AB/CD - PB/CD + PB/CD = AN/NC + BM/MD = > AN/NC + BM/MD = 1; AN = AO - ON si BM = BD - MD = >
(AO - ON)/NC + (BD - MD)/MD = 1 = > AO/NC - ON/NC + BD/MD - MD/MD = 1;
AO=NC - ON; BD = 2 DO = 2.(MD - OM) = 2.MD - 2.OM
Se obtine:
(NC - ON)/NC - ON/NC + (2.MD - 2 .OM)/MD - 1 = 1 = > NC/NC - ON/NC - ON/NC +(2MD)/MD - 2.(OM/MD) = 2 = > 1 - 2.(ON/NC) + 2 - 2. (OM/MD) = 2 = >
-2(ON/NC + OM/MD) = -1 = > 2(ON/NC + OM/MD) = 1 = > ON/NC + OM/MD = 1/2.
P.S. Am considerat P intre A si B.
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