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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1188.
SAEDB=SAJHB, SCFGB=SCKHB,Draw AP perpendicular to CK, draw UV//AP (from B)meet etended AJ,CKSAJKC=K AP, SCKHB=CK BV, SAJHB=AJ BU => SAJKC=SAJHB+SCKHB
Is S1=(AEJ)=(BHD) and S4=(CKF)=(BHG) ,then (ABC)=(JKH) (AJ=BH=CK and paralleles ) ,terefore S+(BLM)=S1+S2+S3+S4+(BLM) or S=S1+S2+S3+S4
Because ABHJ and ABDE are both parellelograms easily Tr. AEJ is congruent to Tr. BDH SSS. So S(BDH) = S1Similarly S(BGH) = S4Now since ABHJ is a parellelogram AB = JH and similarly BC = HK and AC = JK since ACKH is also a parellelogram ( AJ//KC & AJ = BH = CK) So Tr.s ABC and HJK are congruent SSS and so S(ABC) = S(HJK) Subtracting the area of the common triangle BLM,S = S1+S2+S3+S4Sumith PeirisMoratuwaSri Lanka
ACKH .....replace with ACKJ