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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 1187.
http://s22.postimg.org/702hbxx4x/pro_1187.pngLet A’, B’, F’, C’ are projections of A, B, F, C over KHNote that tri. FKH is similar to ABD … ( case SSS) with factor of ½So S(FKH)= ¼. S(ABD)Since FH//AD =S(JHK)=S(AKH)Note that S(EFGHK) is not depend on positions of E and G on BA and BDLet E and G coincide to B we have S(EFGHK)= S(BKH)F is the midpoint of BC => FF’= ½.(BB’+CC’)Yellow area= S(BHK)+S(AKH)= ½.KH.(BB’+AA’)=1/2. KH.(BB’+CC’)= KH. FF’=2. S(FKH)= ½. S(ABD)
Is (EBF)=1/2.(BEC) ,(AEK)=1/2.(AEC) ,or(BEF)+(AEK)=1/2(ABC) and (GFCD)=1/2(BCD). Therefore 2.(EFGHK)=2.(ABCD)-2.(EBF)-2.(AEK)-2.(AKJ)--2.(JDH)-2.(GFCD)=(ABD)+(BCD)+(ABC)+(ACD)-(ABC)-(ACD)-(BCD)=(ABD)
Peter - your statement Note that S(EFGHK) is not depend on positions of E and G on BA and BDCan u pls explain how u arrived at this?
Note that S(EFGHK)= S(EFK)+S(KFH)- S(FGH)for any position of E on AB , the distance from E to line FK is an invariant so S(EFK) is an invariant.Similarly for any position of G on BD, S(FGH) is an invariant.S(KFH) is not depend on position of E or G.so S(EFGHK) is not depend on positions of E and G on BA and BD
Let S(BEF) = S(CEF) = PS(AEK) = S(CEK) = Q,S(AJK) = S(CJK) = R,S(DJH) = S(CJH) = U,S(BFG) = S(CFG) = V andS(CGH) = S(DGH) = WS(EFFHJK) = S(ABCD) - P-Q-R-U-V-W ...(1)Now P+Q+R+U+V+W = 1/2 S(ABC) + 1/2 S(ACD) + 1/2S(BCD) = 1/2 S(ABCD) + 1/2 S(BCD) So from (1) S(EFGHJK) = S(ABCD) -1/2S(ABCD) - 1/2S(BCD) = 1/2S(ABD) Sumith PeirisMoratuwaSri Lanka