Challenging Geometry Puzzle: Problem 1587. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.
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ReplyDeleteDefine point K per sketch
S(ABCD)= 36 => h= 8 ( see sketch)
And GB/GF=2/3
Apply Ceva theorem with point H in triangle ABF
AE/EF x GF/GB x KB/KA=1
Replace values of AE/EF and GF/GB in above we get KA=KB
Apply Van aubel’s theorem for point H in triangle ABF
BH/HE= KB/KA + GB/GF
We get BH/BE= ⅝
S(BEF)= ½. 6.8= 24
S(BGH)/S(BEF)= (BH. BG)/(BE.BF)= ⅝ x ⅖= ¼
So S(BGH)= 6
triangle BCG ~ triangle FDG
ReplyDeleteBC:FD=2:3 → put S(BCG)=4k , S(FDG)=9k
FD:DA=3:7→S(ADG)=21k
FG:GB=3:2→S(GBA)=20k
∴S(ABCD)=S(BCG)+S(ADG)+S(GBA)=4k+21k+20k=45k
45k=36
k=4/5
S(GBA)=16
menelauss theorem (AH/HG)*(2/5)*(6/4)=1
∴AH:HG=5:3
S(BGH)=(HG/AG)*S(ABG)=(3/8)*16=6